Question 5

1. Find the direction of line $L_1$: The direction vector $\vec{v}$ is the cross product of the normal vectors of the two planes: $\vec{v} = (2, 3, 1) \times (1, 2, 1) = (1, -1, 1)$.
2. Equation of $L_2$: Since $L_2$ passes through $P(2, -1, 3)$ and is parallel to $\vec{v}$, its equation is $\frac{x-2}{1} = \frac{y+1}{-1} = \frac{z-3}{1} = \lambda$. A general point $Q$ on $L_2$ is $(\lambda+2, -\lambda-1, \lambda+3)$.
3. Find $Q$: Substitute $Q$ into the plane $M: 2x + y - 2z = 6$. $2(\lambda+2) + (-\lambda-1) - 2(\lambda+3) = 6 \implies -\lambda - 3 = 6 \implies \lambda = -9$. Thus $Q = (-7, 8, -6)$.
4. Calculate $PQ$: $PQ = \sqrt{(-7-2)^2 + (8+1)^2 + (-6-3)^2} = \sqrt{81 + 81 + 81} = 9\sqrt{3}$. Option (A) is correct.
5. Find $R$ (foot of perpendicular from $P$ to $M$): The distance $PR = \frac{|2(2) + 1(-1) - 2(3) - 6|}{\sqrt{2^2 + 1^2 + (-2)^2}} = \frac{|-9|}{3} = 3$. Coordinates of $R$ are found to be $(4, 0, 1)$. Using the foot of perpendicular formula

$R = P - \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}(a,b,c)$

Here $P(2,-1,3)$ and $(a,b,c) = (2,1,-2)$.

$\frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2} = \frac{2(2) + 1(-1) -2(3) -6}{2^2 + 1^2 + (-2)^2} = \frac{-9}{9} = 1$

Thus

$R = (2,-1,3) - (-1)(2,1,-2)$

$R = (2,-1,3) + (2,1,-2)$

$R = (4,0,1)$

6. Calculate $QR$: $QR = \sqrt{(4+7)^2 + (0-8)^2 + (1+6)^2} = \sqrt{121 + 64 + 49} = \sqrt{234}$. Option (B) is incorrect.
7. Area of $\Delta PQR$: Since $PR \perp QR$, Area $= \frac{1}{2} \times PR \times QR = \frac{1}{2} \times 3 \times \sqrt{234} = \frac{3}{2}\sqrt{234}$. Option (C) is correct.
8. Angle $\theta$ between $PQ$ and $PR$: $\cos \theta = \frac{PR}{PQ} = \frac{3}{9\sqrt{3}} = \frac{1}{3\sqrt{3}}$. Option (D) is incorrect.