Let $$\gamma \in \mathbb{R}$$ be such that the lines $$L_1 : \frac{x+11}{1} = \frac{y+21}{2} = \frac{z+29}{3}$$ and $$L_2 : \frac{x+16}{3} = \frac{y+11}{2} = \frac{z+4}{\gamma}$$ intersect. Let $$R_1$$ be the point of intersection of $$L_1$$ and $$L_2$$. Let $$O = (0, 0, 0)$$, and $$\hat{n}$$ denote a unit normal vector to the plane containing both the lines $$L_1$$ and $$L_2$$.

Match each entry in List-I to the correct entry in List-II.

	List-I		List-II
(P)	$$\gamma$$ equals	(1)	$$-\hat{i} - \hat{j} + \hat{k}$$
(Q)	A possible choice for $$\hat{n}$$ is	(2)	$$\sqrt{\frac{3}{2}}$$
(R)	$$\overrightarrow{OR_1}$$ equals	(3)	1
(S)	A possible value of $$\overrightarrow{OR_1} \cdot \hat{n}$$ is	(4)	$$\frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k}$$
		(5)	$$\sqrt{\frac{2}{3}}$$

Direction ratios of $$L_1$$ are $$\langle 1,2,3\rangle$$ and those of $$L_2$$ are $$\langle 3,2,\gamma\rangle$$.
Write the two lines in the symmetric-to-parametric form.

For $$L_1$$ let the parameter be $$\lambda$$:
$$x=-11+\lambda,\;y=-21+2\lambda,\;z=-29+3\lambda$$

For $$L_2$$ let the parameter be $$\mu$$:
$$x=-16+3\mu,\;y=-11+2\mu,\;z=-4+\gamma\mu$$

Because the lines intersect, there exist $$\lambda,\mu$$ for which the coordinates coincide:

$$-11+\lambda=-16+3\mu\;-(1)$$
$$-21+2\lambda=-11+2\mu\;-(2)$$
$$-29+3\lambda=-4+\gamma\mu\;-(3)$$

Solve $$(1)$$ and $$(2)$$ first.
From $$(2):\;2\lambda-2\mu=10\Rightarrow\lambda-\mu=5\;-(4)$$
From $$(1):\;\lambda-3\mu=-5\;-(5)$$
Substitute $$\lambda=\mu+5$$ from $$(4)$$ into $$(5)$$:
$$\mu+5-3\mu=-5\;\Longrightarrow\;-2\mu=-10\;\Longrightarrow\;\mu=5$$
Hence $$\lambda=\mu+5=10$$.

Coordinates of the point of intersection $$R_1$$ (use $$L_1$$):
$$x=-11+10=-1,\;y=-21+20=-1,\;z=-29+30=1$$
Therefore $$\overrightarrow{OR_1}=\langle-1,-1,1\rangle$$.

Determine $$\gamma$$ using $$(3)$$:
Left side: $$-29+3\lambda=-29+30=1$$
Right side: $$-4+\gamma\mu=-4+5\gamma$$
Equating: $$1=-4+5\gamma\;\Longrightarrow\;5\gamma=5\;\Longrightarrow\;\gamma=1$$.

Now find a normal to the plane containing the two lines.
Two non-parallel direction vectors are
$$\mathbf{v}_1=\langle1,2,3\rangle,\qquad\mathbf{v}_2=\langle3,2,1\rangle$$ (because $$\gamma=1$$).
Take the cross product:

$$\mathbf{n}_0=\mathbf{v}_1\times\mathbf{v}_2 =\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 1&2&3\\ 3&2&1 \end{vmatrix} =\langle-4,\,8,\,-4\rangle=-4\langle1,-2,1\rangle$$

An associated unit normal vector is therefore
$$\hat{n}=\frac{1}{\sqrt6}\langle1,-2,1\rangle =\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}$$.

Finally, compute the scalar product $$\overrightarrow{OR_1}\cdot\hat{n}$$:

$$\langle-1,-1,1\rangle\cdot\frac{1}{\sqrt6}\langle1,-2,1\rangle =\frac{1}{\sqrt6}\,[-1(1)+(-1)(-2)+1(1)] =\frac{1}{\sqrt6}\,( -1+2+1)=\frac{2}{\sqrt6} =\sqrt{\frac{2}{3}}$$

Summarising our results:
$$\gamma=1,\quad\overrightarrow{OR_1}=\,-\hat{i}-\hat{j}+\hat{k},\quad \hat{n}=\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k},\quad \overrightarrow{OR_1}\cdot\hat{n}=\sqrt{\frac{2}{3}}$$

Matching with List-II:
(P) $$\gamma$$ → (3) $$1$$
(Q) $$\hat{n}$$ → (4) $$\frac{1}{\sqrt{6}}\hat{i}-\frac{2}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k}$$
(R) $$\overrightarrow{OR_1}$$ → (1) $$-\hat{i}-\hat{j}+\hat{k}$$
(S) $$\overrightarrow{OR_1}\cdot\hat{n}$$ → (5) $$\sqrt{\frac{2}{3}}$$

Thus the correct option is
Option C: (P) → (3), (Q) → (4), (R) → (1), (S) → (5).