Question 16

Any point on $L_1$ can be expressed as $P_1(\lambda) = (\lambda - 11, 2\lambda - 21, 3\lambda - 29)$. Any point on $L_2$ can be expressed as $P_2(\mu) = (3\mu - 16, 2\mu - 11, \gamma\mu - 4)$. Since the lines intersect, for some $\lambda$ and $\mu$: $\lambda - 11 = 3\mu - 16 \Rightarrow \lambda - 3\mu = -5$ ...(i) $2\lambda - 21 = 2\mu - 11 \Rightarrow 2\lambda - 2\mu = 10 \Rightarrow \lambda - \mu = 5$ ...(ii) Subtracting (i) from (ii): $2\mu = 10 \Rightarrow \mu = 5$. From (ii), $\lambda - 5 = 5 \Rightarrow \lambda = 10$. Equating $z$-coordinates: $3\lambda - 29 = \gamma\mu - 4 \Rightarrow 3(10) - 29 = 5\gamma - 4 \Rightarrow 1 = 5\gamma - 4 \Rightarrow \gamma = 1$. Thus, (P) matches (3).

The point of intersection $R_1$ is $P_1(10) = (10-11, 2(10)-21, 3(10)-29) = (-1, -1, 1)$. $\vec{OR_1} = -\hat{i} - \hat{j} + \hat{k}$, so (R) matches (1).

The direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{v_2} = 3\hat{i} + 2\hat{j} + \hat{k}$. The normal vector to the plane is $\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{vmatrix} = \hat{i}(2-6) - \hat{j}(1-9) + \hat{k}(2-6) = -4\hat{i} + 8\hat{j} - 4\hat{k}$. Magnitude $|\vec{n}| = \sqrt{(-4)^2 + 8^2 + (-4)^2} = \sqrt{16 + 64 + 16} = \sqrt{96} = 4\sqrt{6}$. Unit normal $\hat{n} = \pm \frac{1}{4\sqrt{6}}(-4\hat{i} + 8\hat{j} - 4\hat{k}) = \pm \frac{1}{\sqrt{6}}(-\hat{i} + 2\hat{j} - \hat{k}) = \mp (\frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k})$. A possible choice for $\hat{n}$ is $\frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k}$, so (Q) matches (4).

$\vec{OR_1} \cdot \hat{n} = (-1, -1, 1) \cdot \pm (\frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}) = \pm (\frac{1 - 2 - 1}{\sqrt{6}}) = \pm (-\frac{2}{\sqrt{6}}) = \mp \sqrt{\frac{4}{6}} = \mp \sqrt{\frac{2}{3}}$. A possible value is $\sqrt{2/3}$, so (S) matches (5).

Correct Option: (C)