Let $\mathrm{\gamma <!--\; \gamma \; -->}\in <!--\; \in \; -->\mathbb{R}$be such that the lines $L_1:\frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}$ and $L_2:\frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\mathrm{\gamma <!--\; \gamma \; -->}}$ intersect. Let 1R be the point of intersection of L₁ and L₂. Let $O=(0,0,0),$and $\stackrel{^<!--\; ^\; -->}{n}$ denote a unit normal vector to the plane containing both the lines L₁ and L₂.

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

List-I	List-II
(P) γ equals	(1) $-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{i}-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{j}+\stackrel{^<!--\; ^\; -->}{k}$
(Q) A possible choice for $\stackrel{^<!--\; ^\; -->}{n}$ is	(2) $\sqrt{\frac{3}{2}}$
(R) ${\stackrel{\to <!--\; \to \; -->}{OR}}_1$ equals	(3) 1
(S) A possible value of ${\stackrel{\to <!--\; \to \; -->}{OR}}_1.$ $\stackrel{^<!--\; ^\; -->}{n}$ is	(4) $\frac{1}{\sqrt{6}}\stackrel{^<!--\; ^\; -->}{i}-<!--\; -\; -->\frac{2}{\sqrt{6}}\stackrel{^<!--\; ^\; -->}{j}+\frac{1}{\sqrt{6}}\stackrel{^<!--\; ^\; -->}{k}$
	(5) $\sqrt{\frac{2}{3}}$

The correct option is

The correct match is (P) → (3), (Q) → (4), (R) → (1), (S) → (5).

Step-by-step Derivation:

1. Finding the Point of Intersection and γ:
Let the symmetric equations of the lines $L_1$ and $L_2$ be equated to parameters $\lambda$ and $\mu$ respectively:

$L_1:\frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}=\lambda$

$L_2:\frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma }=\mu$

Any general point on $L_1$ can be expressed as:
$(\lambda -11,2\lambda -21,3\lambda -29)$

Similarly, any general point on $L_2$ can be expressed as:
$(3\mu -16,2\mu -11,\gamma \mu -4)$

Since the lines intersect at a point $R$, their coordinates must be equal for some values of $\lambda$ and $\mu$:
From the x-coordinates:
$\lambda -11=3\mu -16\Rightarrow \lambda -3\mu =-5$ --- (Equation 1)
From the y-coordinates:
$2\lambda -21=2\mu -11\Rightarrow 2\lambda -2\mu =10\Rightarrow \lambda -\mu =5$ --- (Equation 2)

Subtracting Equation 1 from Equation 2 gives:
$2\mu =10\Rightarrow \mu =5$
Substituting $\mu =5$ into Equation 2:
$\lambda =10$

Using these parameter values, the coordinates of the point of intersection $R$ are:
$x=10-11=-1$
$y=2\left(10\right)-21=-1$
$z=3\left(10\right)-29=1$
Thus, the position vector $\overrightarrow{OR}$ is:

$\overrightarrow{OR}=-\hat{i}-\hat{j}+\hat{k}$

This matches entry (1) in List-II, meaning (R) → (1).

Now, equating the z-coordinates at the intersection point:
$z=\gamma \mu -4\Rightarrow 1=5\gamma -4\Rightarrow 5\gamma =5\Rightarrow \gamma =1$
This matches entry (3) in List-II, meaning (P) → (3).

2. Finding the Unit Normal Vector $\hat{n}$:
The direction vector of line $L_1$ is $\overrightarrow{d_1}=\hat{i}+2\hat{j}+3\hat{k}$.
The direction vector of line $L_2$ (with $\gamma =1$) is $\overrightarrow{d_2}=3\hat{i}+2\hat{j}+\hat{k}$.
The normal to the plane containing both lines is parallel to the cross product of these two directions:

$\overrightarrow{d_1}\times \overrightarrow{d_2}=\left|\begin{array}{c}\hat{i}& \hat{j}\hat{k}\\ 123321\end{array}\right|$

$=\hat{i}(2-6)-\hat{j}(1-9)+\hat{k}(2-6)=-4\hat{i}+8\hat{j}-4\hat{k}$

We can simplify the direction vector of the normal to $\hat{i}-2\hat{j}+\hat{k}$.
The unit normal vector $\hat{n}$ is:

$\hat{n}=\pm \frac{\hat{i}-2\hat{j}+\hat{k}}{\sqrt{1^2+{(-2)}^2+1^2}}=\pm (\frac{1}{\sqrt{6}}\hat{i}-\frac{2}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k})$

This matches entry (4) in List-II, meaning (Q) → (4).

3. Calculating $\overrightarrow{OR}\cdot \hat{n}$:
Using $\overrightarrow{OR}=-\hat{i}-\hat{j}+\hat{k}$ and choosing the unit normal vector $\hat{n}=\frac{1}{\sqrt{6}}\hat{i}-\frac{2}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k}$:

$\overrightarrow{OR}\cdot \hat{n}=(-1)\left(\frac{1}{\sqrt{6}}\right)+(-1)(-\frac{2}{\sqrt{6}})+\left(1\right)\left(\frac{1}{\sqrt{6}}\right)$

$=-\frac{1}{\sqrt{6}}+\frac{2}{\sqrt{6}}+\frac{1}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{4}{6}}=\sqrt{\frac{2}{3}}$

This matches entry (5) in List-II, meaning (S) → (5).