Let \(\gamma \in \mathbb{R}\) be such that the lines \(\mathrm{L}_1: \frac{x+11}{1}=\frac{\mathrm{y}+21}{2}=\frac{\mathrm{z}+29}{3}\) and \(L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}\) intersect. Let \(\mathrm{R}_1\) be the point of intersection of \(\mathrm{L}_1\) and \(\mathrm{L}_2\). Let \(\mathrm{O}=(0,0,0)\), and \(\hat{\mathrm{n}}\) denote a unit normal vector to the plane containing both the lines \(\mathrm{L}_1\) and \(\mathrm{L}_2\).
Match each entry in List-I to the correct entry in List-II.

The correct option is

\(\begin{aligned} & \mathrm{L}_1: \frac{\mathrm{x}+11}{1}=\frac{\mathrm{y}+21}{2}=\frac{\mathrm{z}+29}{3}=\mathrm{a} \\ & \mathrm{L}_2: \frac{\mathrm{x}+16}{3}=\frac{\mathrm{y}+11}{2}=\frac{\mathrm{z}+4}{\gamma}=\mathrm{b}\end{aligned}\)
\(\begin{aligned} & x=a-11=3 b-16 \Rightarrow a-3 b=-5...(1) \\ & y=2 a-21=2 b-11 \Rightarrow 2 a-2 b=10...(2) \\ & z=3 a-29=b r-4 \Rightarrow 3 a-b \gamma=25...(3)\end{aligned}\)
from (1) & (2)
\(\mathrm{a}=10, \mathrm{~b}=5\)
Now from (3)
\(3(10)-5 \gamma=25 \quad \therefore \gamma=1\)
\(\begin{aligned} & \mathrm{R}_1 \equiv(-1,-1,1) \\ & \mathrm{OR}_1=-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{n}}=\left|\begin{array}{lrr}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 2 & 3 \\ 3 & 2 & 1\end{array}\right|=-4 \hat{\mathrm{i}}-(-8) \hat{\mathrm{j}}-4 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{n}}=-4 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}=-4(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & \hat{\mathrm{n}}= \pm \frac{4(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{4 \sqrt{6}}= \pm \frac{(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{\sqrt{6}} \\ & \overrightarrow{\mathrm{OR}}_1 \cdot \hat{\mathrm{n}}= \pm(-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\left(\frac{\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{6}}\right)= \pm \frac{2}{\sqrt{6}}= \pm \sqrt{\frac{4}{6}}= \pm \sqrt{\frac{2}{3}}\end{aligned}\)