Intersection of Lines and Normal Vector to Plane

To find the correct matching between List-I and List-II, we analyze the given lines and perform the required step-by-step vector and 3D geometry computations.

Step 1: Finding $\gamma$ and the point of intersection $R_1$

The given lines are: $L_1 : \frac{x+11}{1} = \frac{y+21}{2} = \frac{z+29}{3} = t_1$ $L_2 : \frac{x+16}{3} = \frac{y+11}{2} = \frac{z+4}{\gamma} = t_2$

Any general point on line $L_1$ is given by: $(x, y, z) = (t_1 - 11, 2t_1 - 21, 3t_1 - 29)$

Any general point on line $L_2$ is given by: $(x, y, z) = (3t_2 - 16, 2t_2 - 11, \gamma t_2 - 4)$

Since $L_1$ and $L_2$ intersect at $R_1$, we equate their $x$ and $y$ coordinates:

1. $t_1 - 11 = 3t_2 - 16 \implies t_1 - 3t_2 = -5 \quad \text{--- (1)}$
2. $2t_1 - 21 = 2t_2 - 11 \implies 2t_1 - 2t_2 = 10 \implies t_1 - t_2 = 5 \quad \text{--- (2)}$

Subtracting equation (1) from equation (2): $(t_1 - t_2) - (t_1 - 3t_2) = 5 - (-5)$ $2t_2 = 10 \implies t_2 = 5$

Substituting $t_2 = 5$ into equation (2): $t_1 - 5 = 5 \implies t_1 = 10$

Now, substituting $t_1 = 10$ and $t_2 = 5$ into the $z$-coordinates: $z = 3(10) - 29 = 1$ $z = 5\gamma - 4$

Equating the $z$-coordinates: $5\gamma - 4 = 1 \implies 5\gamma = 5 \implies \gamma = 1$

Thus, (P) $\rightarrow$ (3).

Substituting $t_1 = 10$ into the parametric equation of $L_1$, the point of intersection $R_1$ is: $R_1 = (10 - 11, 2(10) - 21, 3(10) - 29) = (-1, -1, 1)$

Therefore, the position vector $\overrightarrow{OR_1}$ is: $\overrightarrow{OR_1} = -\hat{i} - \hat{j} + \hat{k}$

Thus, (R) $\rightarrow$ (1).

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Step 2: Finding a possible unit normal vector $\hat{n}$

The direction vectors of lines $L_1$ and $L_2$ are: $\vec{d}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ $\vec{d}_2 = 3\hat{i} + 2\hat{j} + \hat{k} \quad (\text{since } \gamma = 1)$

A normal vector $\vec{n}$ to the plane containing both lines is given by the cross product $\vec{d}_1 \times \vec{d}_2$: $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(1 - 9) + \hat{k}(2 - 6) = -4\hat{i} + 8\hat{j} - 4\hat{k}$

Dividing by $-4$, a vector in the direction of the normal is $\hat{i} - 2\hat{j} + \hat{k}$. The unit normal vector $\hat{n}$ is: $\hat{n} = \pm \frac{\hat{i} - 2\hat{j} + \hat{k}}{\sqrt{1^2 + (-2)^2 + 1^2}} = \pm \left( \frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k} \right)$

A possible choice for $\hat{n}$ is: $\hat{n} = \frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k}$

Thus, (Q) $\rightarrow$ (4).

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Step 3: Calculating $\overrightarrow{OR_1} \cdot \hat{n}$

Using $\overrightarrow{OR_1} = -\hat{i} - \hat{j} + \hat{k}$ and $\hat{n} = \frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k}$: $\overrightarrow{OR_1} \cdot \hat{n} = (-\hat{i} - \hat{j} + \hat{k}) \cdot \left( \frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k} \right)$ $\overrightarrow{OR_1} \cdot \hat{n} = \frac{1}{\sqrt{6}} \left[ (-1)(1) + (-1)(-2) + (1)(1) \right] = \frac{1}{\sqrt{6}} [-1 + 2 + 1] = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}$

Thus, (S) $\rightarrow$ (5).

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Conclusion

The correct matching is:

• (P) $\rightarrow$ (3)
• (Q) $\rightarrow$ (4)
• (R) $\rightarrow$ (1)
• (S) $\rightarrow$ (5)

This corresponds to Option C.