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.

\textbf{List-I}

(P) $\gamma$ equals

(Q) A possible choice for $\hat { n }$ is

(R) $\overrightarrow { O R _ { 1 } }$ equals

(S) A possible value of $\overrightarrow { O R _ { 1 } } \cdot \hat { n }$ is

\textbf{List-II}

(1) $- \hat { i } - \hat { j } + \hat { k }$

(2) $\sqrt { \frac { 3 } { 2 } }$

(3) 1

(4) $\frac { 1 } { \sqrt { 6 } } \hat { i } - \frac { 2 } { \sqrt { 6 } } \hat { j } + \frac { 1 } { \sqrt { 6 } } \hat { k }$

(5) $\sqrt { \frac { 2 } { 3 } }$

The correct option is

(A) $(\mathrm{P}) \rightarrow (3)$, $(\mathrm{Q}) \rightarrow (4)$, $(\mathrm{R}) \rightarrow (1)$, $(\mathrm{S}) \rightarrow (2)$

(B) $(\mathrm{P}) \rightarrow (5)$, $(\mathrm{Q}) \rightarrow (4)$, $(\mathrm{R}) \rightarrow (1)$, $(\mathrm{S}) \rightarrow (2)$

(C) $(\mathrm{P}) \rightarrow (3)$, $(\mathrm{Q}) \rightarrow (4)$, $(\mathrm{R}) \rightarrow (1)$, $(\mathrm{S}) \rightarrow (5)$

(D) $(\mathrm{P}) \rightarrow (3)$, $(\mathrm{Q}) \rightarrow (1)$, $(\mathrm{R}) \rightarrow (4)$, $(\mathrm{S}) \rightarrow (5)$