1. \(2x + y + z = k\)

\(x = 0, y = 0 \Rightarrow x - axis\)

\(y = 0, z = 0 \Rightarrow x - axis\)

\(x = 0, z = 0 \Rightarrow y - axis\)

volume of tetrahedron \(= \frac{1}{6} [\overline{AB} \overline{AC} \overline{AD}]\)

\(V = \frac{1}{6} \left| \begin{array}{ccc} 0 & k & 0 \\ 0 & 0 & k \\ -k & 0 & 0 \end{array} \right| = \frac{1}{6} \frac{k^{3}}{2}\)

\(\frac{2V^{3}}{3} = \frac{1}{6} \frac{k^{3}}{2}\)

\(V^{3} : k^{3} = 1^{3} : 2^{3} \Rightarrow k : V = 2 : 1\)

2. d.r's of the plane \(\pi_{1} = 1, -2, 5\)

d.r's of the plane \(\pi_{2} = 0, 4, - 6\)

let \(\theta\) be the angle between \(\pi_{1}\) and \(\pi_{2}\)

3. Given plane \(2x - y - 2z - 9 = 0\) Distance from \(O(0,0,0)\) to plane \(ax + by + cz + d = 0\) is \(distance = \frac{|d|}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{|-9|}{\sqrt{4 + 1 + 4}} = 1\)

4. \(\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 1}{-1} = t\)

\(P(x,y,z) = (2t + 1, t - 1, - t + 1)\)

\(x + 2y + 3z = 4\)

\((2t + 1) + 2(t - 1) + 3(t - 1) = 4 \Rightarrow t = 2\)

\(P(5,1, - 1),\)

\((2\vec{i} - 3\vec{j}) \times (\vec{i} + 2\vec{j} - \vec{k}) = 3\vec{i} + 2\vec{j} + 7\vec{k}\)

Req line is

\(\frac{x - 5}{3} = \frac{y - 1}{2} = \frac{z + 1}{7} (or)\)

\(\frac{x - 5}{-3} = \frac{y - 1}{-2} = \frac{z + 1}{-7}\)

5. \(\pi_{1} + \lambda \pi_{2} = 0\) passes through the point \((0,0,0) \Rightarrow \lambda = 4\) required plane is \(x + 2y + 3z - 4 + 4(4x + 3y + 2z + 1) = 0 \Rightarrow 17x + 14y + 11z = 0\)

6. \((a_{1}, b_{1}, c_{1})\) are dr's of \(1^{st}\) line

\(\frac{a_{1}}{-1} = \frac{b_{1}}{-1} = \frac{c_{1}}{3}\)

\((a_{1}, b_{1}, c_{1}) = (-1, - 1, 3)\)

\((a_{2}, b_{2}, c_{2})\) are dr's of \(2^{nd}\) line.

\(\frac{a_{2}}{1} = \frac{b_{2}}{5} = \frac{c_{2}}{2} (a_{2}, b_{2}, c_{2}) = (1, 5, 2)\)

Since \(a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0 \Rightarrow \theta = 90^{\circ}\)

7. Let \(\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 5}{-3} = r \rightarrow (1) \&\)

\(\frac{x + 5}{3} = \frac{y - 4}{-1} = \frac{z + 3}{4} = s \rightarrow (2)\)

\((x_{1}, y_{1}, z_{1}) = (r + 1, 2r + 2, - 3r + 5)\)

\((x_{2}, y_{2}, z_{2}) = (3s - 5, - s + 4, 4s - 3)\)

\(r + 1 = 3s - 5\)

\(3s - r - 6 = 0 \rightarrow (3)\)

\(2r + 2 = -s + 4\)

\(s + 2r - 2 = 0 \rightarrow (4)\)

\(\frac{s}{2 + 12} = \frac{1}{6 + 1} \Rightarrow s = 2\)

\(z = 4s - 3 = 4(2) - 3 = 5\)

\(\frac{x + 1}{-1} = \frac{y - 3}{\sin 120^{\circ}} = r \quad (8)\)

9. \(\cos \theta = \frac{3 + 6 + 4}{\sqrt{1 + 4 + 4}\sqrt{9 + 9 + 4}} = \frac{13}{3\sqrt{22}}\)

10. Conceptual

11. given \(\frac{x}{1} = \frac{y}{2} = \frac{z}{3} \dots (1)\)

\(\frac{x + 2}{-3} = \frac{y - 1}{2} = \frac{z + 1}{5} \dots (2)\)

let \(a, b, c\) are dr's of req.line \(\perp\) rto \((1) \& (2)\)

\(a + 2b + 3c = 0 \& -3a + 2b + 5c = 0\)

by solving weget \(\frac{a}{2} = \frac{b}{- 7} = \frac{c}{4}\)

\(req.line through (-1, 3, - 2) is \frac{x + 1}{2} = \frac{y - 3}{-7} = \frac{z + 2}{4}\)

12. Equation of line \(\frac{x - 1}{2} = \frac{y - 1}{-2} = \frac{z + 1}{1}\) \(\tan \theta = \frac{1}{\sqrt{8}} \Rightarrow \sin \theta = \frac{1}{3}\) \(\sin \theta = \frac{al + bm + cn}{\sqrt{a^{2} + b^{2} + c^{2}}\sqrt{l^{2} + m^{2} + n^{2}}}\) \(\frac{1}{3} = \frac{2\sqrt{\lambda} - 6 + 6}{\sqrt{9\lambda + 9 + 36}} \Rightarrow \lambda = 15\)

13. \(2x^{2} - 6y^{2} - 12z^{2} + 18yz + 2zx + xy\) \(= (x + 2y - 2z + k)(2x - 3y + 6z + l)\) \(\cos \theta = \frac{|a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2}|}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}}\sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}\) \(\cos \theta = \frac{16}{21} \Rightarrow \theta = \cos^{-1}\left(\frac{16}{21}\right)\)

14. Normal to the plane \(AB \times AC = (1, -2, 1)\)

\(d_{c's} = \left(\frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)\)

\(\cos \alpha = \frac{1}{\sqrt{6}}, \cos \beta = \frac{2}{\sqrt{6}}\)

\(\sin^{2}\alpha + \sin^{2}\beta = 1 - \frac{1}{6} + 1 - \frac{4}{6} = \frac{7}{6}\)

15. Equation of \(\pi_{1}\) is \(y + 2z - 13 = 0\)

Equation of \(\pi_{2}\) is \(x - 2y + z + 3 = 0\)

Here \(a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0 \Rightarrow \theta = \frac{\pi}{2}\)

16. Equation of the plane passing through line of intersection of planes \(\pi_{1}\) and \(\pi_{2}\) is

\(\pi_{1} + \lambda \pi_{2} = 0\)

\(\Rightarrow (2x + 6y + 4z - 7) + \lambda (x - y - 2z - 2) = 0\)

\(\Rightarrow (2 + \lambda)x + (6 - \lambda)y + (4 - 2\lambda)z - (7 + 2\lambda) = 0\)

Since the above plane is perpendicular to \(x + y + 2z - 5 = 0\)

We can write

\((2 + \lambda) + (6 - \lambda) + 2(4 - 2\lambda) = 0 \Rightarrow \lambda = 4\)

Required equation of plane is

\(6x + 2y - 4z - 15 = 0\)

17. Dr's of normal to the plane are (a,b,1) Dr's of line are (1,1,2) But line is perpendicular to the normal to the plane. Then a+b+2=0 a+b=-2

18. Let the required plane

\(a x + b y + c z + d = 0 \rightarrow (1)\)

\((1) passes through (2,1,3)\)

\(2a + b + 3c + d = 0 \rightarrow (2)\)

\((1) \bot^{r} to x - 2y + 2z + 3 = 0 \& 3x - 2y + 4z - 4 = 0\)

\(a - 2b + 2c = 0 \rightarrow (3)\)

\(\& 3a - 2b + 4c = 0 \rightarrow (4)\)

solving (2),(3),(4)

we get \(a = 2, b = -1, c = -2, d = 3\)

i.e \(2x - y - 2z + 3 = 0\)

19. \(\pi \rightarrow x + y + z = 12\)

\(image of A is = Q(5,6,7)\)

\(\Rightarrow \frac{x - 5}{2} = \frac{y - 6}{1} = \frac{z - 7}{-2} = k\)

\(B(2k + 5, k + 6, -2k + 7)\)

\(\Rightarrow 2k + 5 + k + 6 - 2k + 7 = 12\)

\(k = -6\)

\(OB = \sqrt{49 + 361} = \sqrt{410}\)

20. Given \(4x + 3y + 2z = 2\)

sum of intercepts \(\frac{1}{2} + \frac{2}{3} + 1 = \frac{13}{6}\)

21. \(\pi_{1} : 2x - y + z - 6 = 0\)

\(\pi_{2} : x + y + 2z - 3 = 0\)

\(\cos \theta = \frac{|2 - 1 + 2|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}\)

\(\theta = \frac{\pi}{3}\)

22. Let the plane \(a x + b y + c z + d = 0\)

Passes \((0,1,2)\) \(b + 2c + d = 0 \rightarrow (2)\)

\(passes (-1,0,3) - a + 3c + d = 0 \rightarrow (3)\)

\((1) \bot^{r} to 2x + 3y + z - 5 = 0\)

\(2a + 3b + c = 0 \rightarrow (4)\)

by solving \(a = 4, b = -3, c = 1\)

\(\therefore (1) \rightarrow 4x - 3y + z + 1 = 0\)

23. \(dr's of PQ = (2,1,2)\)

Equation of the plane

\(2(x - 2) + 1(y - 1) + 2(z - 2) = 0\)

\(2x + y + 2z = 9\)

24. \(\alpha = 45^{\circ}, \beta = 120^{\circ}, \gamma = ?\)

\(\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1\)

\(\frac{1}{4} + \frac{1}{4} + \cos^{2}\gamma = 1 \Rightarrow \cos^{2}\gamma = \frac{1}{4} \Rightarrow \gamma = 60^{\circ}\)

25. \(G = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) = (x, y, z)\)

\(a = 3x, b = 3y, c = 3z\)

Given \(\perp^{r}\) distance is 1

\(\Rightarrow \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} = 1 \Rightarrow \frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = 9\)

26. The required plane

\(\Rightarrow 3x - 4z = - 1\) (parallel to y- axis)

27. \(\pi : x + 2y - 2z + 5 = 0\)

\(d = \perp^{r} distance from (0,0,0) to \pi\)

\(d = \frac{5}{\sqrt{1 + 4 + 4}} = \frac{5}{\sqrt{9}} = \frac{5}{3} units\)

28. Required equation is

\((x - 2y + z + 2) + \lambda (3x - y - z + 1) = 0 \dots (1)\)

\(\Rightarrow \lambda = -1\)

Substitute \(\lambda = - 1\) in equation (1)

\(\Rightarrow 2x + y - 2z = 1\)

\(x-intercept = \frac{1}{2}\)

29. \(\frac{3 - 2}{2} = \frac{2 - 3}{3} = \frac{1 - 2}{2} = 0 \Rightarrow \lambda = 4\)

\(\sin^{-1}(\sin 4) + \cos^{-1}(\cos 4) = \pi - 4 + 2\pi - 4 = 3\pi - 8\)

30. Given \(a x + b y + c z + 1 = 0 \dots (1)\)

\(2x - 2y + z = 0 \dots (2)\)

\(x - y + 2z = 4 \dots (3)\)

\((1) \perp (2) \Rightarrow 2a - 2b + c = 0 \dots (4)\)

\((1) \perp (3) \Rightarrow a - b + 2c = 0 \dots (5)\)

(1) is passing through

\((1, - 2,1) \Rightarrow a - 2b + c = -1 \dots (6)\)

Solving (4),(5) and (6), we get

\(a = 1, b = 1 \& c = 0\)

Now \(a + b - c = 1 + 1 - 0 = 2\)

31. All points lies on plane \(2x - 2y + 4z + 5 = 0\)

\(\overline{r}.(2\hat{i} + 3\hat{j} - 4\hat{k}) = 1\)

\(\Rightarrow (x\hat{i} + y\hat{j} + z\hat{k}).(2\hat{i} + 3\hat{j} - 4\hat{k}) = 1\)

\(\Rightarrow 2x + 3y - 4z - 1 = 0 \dots (1)\)

Equation of plane parallel to equation (1) is

\(2x + 3y - 4z + h = 0 \dots (2)\)

Given : distance between (1) & (2) = 2

\(\Rightarrow \frac{|h + 1|}{\sqrt{4 + 9 + 16}} = 2 \Rightarrow h = -1 \pm 2\sqrt{29}\)

Hence required equation is:

\(2x + 3y - 4z = 1 \pm 2\sqrt{29}\)

33. Given points A(1,1,- 1), B(2,- 1,0) & C(- 1,0,2)

Option verification: \(1 + 2 - 2 - 1 = 0\)

\(\frac{x}{20} + \frac{y}{15} + \frac{z}{-12} = 1\)

A(20,0,0), B(0,15,0), C(0,0,- 12) and O(0,0,0).

Volume of tetrahedron OABC is

35. \(P(x, y, z_{1}) = (1,1,1)\)

X- intercept = 5 / 2

\(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\)

\(\frac{x}{5 / 2} + \frac{y}{b} + \frac{z}{c} = 1\)

\(\frac{2}{5} + \frac{1}{b} + \frac{1}{c} = 1\)

\(\frac{1}{b} + \frac{1}{c} = 1 - \frac{2}{5} = \frac{3}{5}\)

Since y- intercept of the plane \(\pi\) is Negative,

Z- Intercept of the plane \(\pi\) is Positive

Y- Intercept = ?

Distance from O(0,0,0) to eq. (1) = \(\frac{5}{7}\)

\(\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}} = \frac{5}{7}\)

\(25\left(\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}\right) = 49\)

\(\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{49}{25}\)

\(\frac{4}{25} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{49}{25}\)

\(\frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{45}{25} = \frac{9}{5}\)

\(\left(\frac{1}{b} + \frac{1}{c}\right)^{2} - 2\frac{1}{bc} = \frac{9}{5}\)

\(\frac{9}{25} - \frac{9}{5} = \frac{2}{bc} \left(\therefore \frac{1}{b} + \frac{1}{c} = -\frac{3}{5} + \frac{6}{5} = \frac{3}{5}\right)\)

\(bc = - 25 / 18\) (satisfied)

\(= -\frac{5}{3} \cdot \frac{5}{6} = -\frac{25}{18} \therefore b = -\frac{5}{3}, c = \frac{5}{6}\)

y- intercept of the plane \(\pi\) is \(- 5 / 3\)

36. D.r (1,2,2)

\(x + 2y + 2z + r = 0 \dots (1)\)

Distance from O(0,0,0) to eq... (1) = \(\frac{1}{3}\)

\(\frac{r}{\sqrt{1 + 4 + 4}} = \frac{1}{3}\)

\(\left(\frac{r}{3}\right) = \frac{1}{3}\)

\(r = \pm 1\)

\(\therefore x + 2y + 2z + 1 = 0\)

\(\therefore \sqrt{p^{2} + q^{2} + r^{2}} = \sqrt{2^{2} + 2^{2} + 1^{2}} = \sqrt{9} = 3\)

37. \(P(x_{1}, y_{1}, z_{1}) = P(1, 0, 1)\) \(2a + 3b - c = 0\) \(a - b + 2c = 0\) 3 -1 2 3 -1 2 1 1 \(\frac{a}{6 - 1} = \frac{b}{- 1 - 4} = \frac{c}{- 2 - 3}\) \(\frac{a}{5} = \frac{c}{- 5} = \frac{c}{- 5}\)

Eq. of plane having \(dr's (1, - 1, 1) \& P(1, 0, 1)\)

\(1(x - 1) - 1(y - 0) - 1(z - 1) = 0\)

\(x - 1 - y - z + 1 = 0\)

\(dr's \quad 1, - 1, - 1\)

\(P(x_{1}, y_{1}, z_{1}) = P(11, 7, 5)\)

\(1(x - 11) - (y - 7) - (z - 5) = 0\)

\(x - 11 - y + 7 - z + 5 = 0\)

\(x - y - z + 1 = 0\)

\(a = 1, b = -1, c = -1, d = 1\)

\(\frac{a}{b} + \frac{b}{d} = \frac{1}{-1} + \frac{1}{-1} = -1 - 1 = -2\)

38. \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\)

\(\frac{x}{-2} + \frac{y}{4 / 3} - \frac{z}{-4 / 5} = 1\)

\(-2x + 3y - 5z = 4\)

\(2x - 3y + 5z + 4 = 0 \dots (1)\)

\(Dr's \quad 2, - 3, 5\)

\(Dc's \frac{2}{\sqrt{38}}, - \frac{3}{\sqrt{38}}, \frac{5}{\sqrt{38}}\)

39. \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \dots (1)\)

(1) Passes through P(1,2,3)

\(\frac{1}{a} + \frac{2}{b} + \frac{3}{c} = 1 \dots (2)\)

\(3x + 4y - 5z = 0 \dots (3)\)

\(\frac{3}{1 / a} = \frac{4}{1 / b} = -\frac{5}{1 / c}\)

\(3a = 4b = -5c = k\)

\(a = k / 3, b = k / 4, c = -k / 5\)

\((3) \Rightarrow \frac{3}{k} + \frac{8}{k} - \frac{15}{k} = 1\)

\(-\frac{4}{k} = 1 \Rightarrow k = -4\)

\(a = -\frac{4}{3}, b = -1 ; c = \frac{4}{5}\)

\(3a + b + 5c = -4 - 1 + 4 = -1\)

40. Dr \((3 + 1, 4 - 3, - 7 - 6)\)

Dr (5, 1, - 13)

\(5(x - 3) + 1(y - 4) - 13(z + 7) = 0\)

\(5x - 15 + y - 4 - 13z - 91 = 0\)

\(5x + y - 13z - 110 = 0\)

\(\frac{5x}{110} + \frac{y}{110} + \frac{z}{110} = 1\)

\(\frac{x}{22} + \frac{y}{110} + \frac{z}{110} = 1\)

\(a + b = 22 + 110 = 132\)

41. Data is not sufficient.

42. G (2,-3,5)

\(A(a, 0,0) B(0, b, 0) C(0, 0, c)\)

\(\frac{a}{3} = 2, \frac{b}{3} = -3, \frac{c}{3} = 5\)

\(\frac{x}{6} + \frac{y}{-9} + \frac{z}{15} = 1\)

\(15x - 10y + 6z - 90 = 0 \dots (1)\)

\(\perp^{r} distance from O(0,0,0) to eq \dots (1)\)

\(= \frac{90}{\sqrt{225 + 100 + 36}} = \frac{90}{\sqrt{361}} = \frac{90}{19}\)

43. The line segment AB divide in the ratio 2:1

\(\left(\frac{6 - 3}{3}, \frac{2 - 2}{3}, \frac{4 + 7}{3}\right)\)

P(1,0,1)

\(2(x - 1) + 1(y - 0) - 3(z - 1) = 0\)

\(2x + y - 3z + 1 = 0\)

\(-2x - y + 3z = 1\)

\(Y - intercept = -1\)

\(\left| \begin{array}{ccc} x-3 & y-4 & z+1 \\ 4 & 2 & -6 \\ 1 & 5 & -6 \end{array} \right| = 0\)

\(x + y + z - 6 = 0\)

\(ax + by + cz - d = 0\)

a=1,b=1,c=1,d=6,

\(a + c = 2\)

\(3(a + c) = 3(2) = 6\)

\(\therefore d = 6\)

\(\therefore 3(a + c) = d\)

45. d.r. 2,-1,5

\(2(x + 1) - 1(y - 3) + 5(z + 2) = 0\)

\(2x + 2 - y + 3 + 5z + 10 = 0\)

\(2x - y + 5z + 15 = 0 \dots (1)\)

\(\perp^{r} distance from O(0,0,0) to eq....(1)\)

\(\frac{15}{\sqrt{4 + 1 + 25}} = \frac{15}{\sqrt{30}}\)

\(= \frac{15}{\sqrt{15}\sqrt{2}} = \sqrt{\frac{15}{2}}\)

46. AB=1,0,4

\(CD = 7, - 1, - 1\)

\(\cos \theta = \frac{|7 + 0 - 4|}{\sqrt{17}\sqrt{49 + 1 + 1}}\)

\(= \frac{3}{\sqrt{17}\sqrt{51}} = \frac{3}{\sqrt{17}\sqrt{17}\sqrt{3}} = \frac{3}{17\sqrt{3}}\)

\(\left| \begin{array}{ccc} x-3 & y+1 & z-4 \\ -1 & 2 & 1 \\ 1 & 3 & 2 \end{array} \right| = 0\)

\((x - 3)(1) + 3(y + 1) - 5(z - 4) = 0\)

\(x - 3 + 3y + 3 - 5z + 20 = 0\)

\(x + 3y - 5z + 20 = 0 \dots (1)\)

Eq.(1) passes through P(2,1,k)

\(2 + 3 - 5k + 20 = 0\)

\(25 - 5k = 0\)

\(\therefore k = 5\)

48. \(6x - 3y + 2z = 6\)

\(\frac{x}{1} + \frac{y}{-2} + \frac{z}{3} = 1\)

\(a = 1, b = -2, c = 3\)

\(dr's \quad 6, - 3, 2\)

\(dc's \quad \frac{6}{7}, \frac{-3}{7}, \frac{2}{7}\)

\(l = \frac{6}{7}, m = \frac{-3}{7}, n = \frac{2}{7}, p = \frac{6}{7}\)

\(|al + bm + cn| = \frac{6}{7} + \frac{6}{7} + \frac{6}{7} = \frac{18}{7}\)

\(3p = 3\left(\frac{6}{7}\right) = \frac{18}{7}\)

\(|al + bm + cn| = 3p\)

49. \(x + y + z - 5 = 0\)

\(A(1,1,1) B(2,2,2)\)

\(\pi_{11} = 1 + 1 + 1 - 5 = -2 \Rightarrow \pi_{22} = 2 + 2 + 2 - 5 = 1\)

\(-\pi_{11} : \pi_{22} = 2 : 1\)

51. \(x(2 - 0) - (y - 1)(-6 - 0) + (z - 2)(12 + 4) = 0\)

\(2x + 6y + 16z - 38 = 0\)

\(x + 3y + 8z - 19 = 0\)

\(d.r's 1, 3, 8\)

\(d.c's \frac{1}{\sqrt{74}}, \frac{3}{\sqrt{74}}, \frac{8}{\sqrt{74}}\)

\(|l| + |m| + |n| = \frac{1 + 3 + 8}{\sqrt{74}}\)

\(= \frac{12}{\sqrt{74}}\)

52. \(\pi_{1} : 2x - y + 2z + 3 = 0\)

\(\pi_{2} : 2x - y + 2z + \frac{\lambda}{2} = 0\)

\(\pi_{3} : 2x - y + 2z + \mu = 0\)

distance between \(\pi_{1} \& \pi_{2}\) is \(\frac{1}{3}\)

\(\Rightarrow \frac{\left|\frac{\lambda}{2} - 3\right|}{\sqrt{4 + 1 + 4}} = \frac{1}{3}\)

\(\Rightarrow \frac{\lambda}{2} - 3 = \pm 1\)

\(\Rightarrow \frac{\lambda}{2} = 4 \quad \frac{\lambda}{2} = 2\)

\(\Rightarrow \lambda = 8 \quad \lambda = 4\)

distance between \(\pi_{1} \& \pi_{3}\) is \(\frac{2}{3}\)

\(\Rightarrow \left|\frac{\mu - 3}{\sqrt{4 + 1 + 4}}\right| = \frac{2}{3}\)

\(\Rightarrow \mu - 3 = \pm 2\)

\(\Rightarrow \mu = 5 \quad \mu = 1\)

Then value of \(\lambda + \mu = 8 + 5 = 13\)

53. Circum centre of right angle \(\Delta le\) is midpoint of hypothesis

Circum centre \(= (2, - 1)\)

And radius \(= \sqrt{5}\)

\(PQ = CP - r\)

\(= \sqrt{(2 + 2)^2 + (-1 + 4)^2} - \sqrt{5}\)

\(= 5 - \sqrt{5}\)

54. Given plane is \(56x + 4y + 9z = 2016\) \(\Rightarrow \frac{x}{36} + \frac{y}{504} + \frac{z}{224} = 1\) \(A = (36,0,0), B = (0,504,0)\) \(C = (0,0,224)\) Centroid of \(\triangle ABC\) \(G = \left(\frac{36}{3}, \frac{504}{3}, \frac{224}{3}\right)\) \(= (12,168,\frac{224}{3})\)

55. \((0,0,1), (0, - 1,0), (\sqrt{2},1,4)\)

Equation of plane

\(-\sqrt{2} x - y + z - 1 = 0\)

By option verification Option (2) -

56. Conceptual

57. Required plane \(x - 2y + 2z + k = 0\)

\(x - 2y + 2z + 3 = 0\)

Required plane is (or) \(x - 2y + 2z - 3 = 0\)

58. let (a,b,c) are Dr's of normal of the required plane \(x - y + 2z - 3 = 0, 2x - 2y + z + 12 = 0\)

\(a - b + 2c = 0 \dots (1)\)

\(2a - 2b + c = 0 \dots (2)\)

Solve (1) and (2)

\(\frac{a}{3} = \frac{b}{3} = \frac{c}{0}\)

Required plane is

\(3(x - 1) + 3(y - 2) = 0\)

\(\Rightarrow x - 1 + y - 2 = 0\)

\(\therefore x + y - 3 = 0\)

59. Required equation is

\(a(x - x_{1}) + b(y - y_{1}) + c(z - z_{1}) = 0\) \((a,b,c) = (1,2,3)\) \((x_{1}, y_{1}, z_{1}) = (1,2,3)\)

60. \(\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1\)

\(\alpha = \beta = \gamma\)

\(3\cos^{2}\alpha = 1 \Rightarrow \cos \alpha = \frac{1}{\sqrt{3}}\)

\(a : b : c = \frac{1}{\sqrt{3}} : \frac{1}{\sqrt{3}} : \frac{1}{\sqrt{3}} = 1 : 1 : 1\)

Eqn of plane is

\(a(x - x1) + b(y - y1) + c(z - z1) = 0\)

\(1(x + 1) + 1(y - 2) + 1(z - 3) = 0\)

\(x + y + z - 4 = 0\)

61. \(\frac{1}{2} = \frac{1}{3} = k = 6\)

\(P(6,6,6)\)

\(dr's (a,b,c) = (5,6,7)\)

\(a(x - x1) + b(y - y1) + c(z - z1) = 0\)

\(5(x - 6) + 6(y - 6) + 7(z - 6) = 0\)

62. Dr's of Normal = Dr's of AB (a,b,c) = (-5,-3,11) = (5,3,-11) midpoint of AB = \(\left(\frac{3}{2}, \frac{7}{2}, \frac{- 9}{2}\right)\) \(x_{1}, y_{1}, z_{1}\) Eqn of the plane is \(a(x - x_{1}) + b(y - y_{1}) + c(z - z_{1}) = 0\)

63. \(\left| \begin{array}{ccc} \bar{i} & \bar{j} & \bar{k} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{array} \right|\) \(= \bar{i}(3) - 3\bar{j} + 3\bar{k}\) dr's of the line are (1,-1,1) dc's of the line are \(\left(\frac{1}{\sqrt{3}}, \frac{- 1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\) \(\therefore \cos \alpha = \frac{1}{\sqrt{3}}\)

64. Point on plane \(\pi\) is (1,-2,1) dr's of plane \(\pi\) are (1,-1,1) = (a,b,c) Equation of \(\pi\) plane is \(a(x - x_{1}) + b(y - y_{1}) + c(z - z_{1}) = 0\) \(x - y - z - 2 = 0\)

\(a = 1, b = -1, c = -1\)

\(b - 2c = -1 - 2(-1) = 1 = a\)

65. Equation of plane is

\(2(x - 2) - 1(y + 1) + 3(z - 3) = 0\) \(2x - y + 3z - 14 = 0\)

66. \(A(6,3,2), B(1, - 4, - 9)\)

d. r of AB \(= 5,7,11\)

\(P(x_{1}, y_{1}, z_{1}) = P(1,1,1)\)

Equation of the plane

\(5(x - 1) + 7(y - 1) + 11(z - 1) = 0\)

\(a = 5, b = 7, c = 11\)

\(\therefore a + b - c = 12 - 11 = 1\)