Let gamma in R be such that the lines L_1: fr | vedclass

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(C) Given lines are $L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3} = a$ and $L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma} = b$.For $L_1$,

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

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(C) Given lines are $L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3} = a$ and $L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma} = b$.For $L_1$,$x = a-11, y = 2a-21, z = 3a-29$.For $L_2$,$x = 3b-16, y = 2b-11, z = b\gamma-4$.Equating coordinates for intersection:$a-11 = 3b-16 \Rightarrow a-3b = -5$ (Eq. $1$)$2a-21 = 2b-11 \Rightarrow 2a-2b = 10 \Rightarrow a-b = 5$ (Eq. $2$)Solving $(1)$ and $(2)$: $a=10, b=5$.Substitute into $z$ coordinates: $3(10)-29 = 5\gamma-4 \Rightarrow 1 = 5\gamma-4 \Rightarrow 5\gamma = 5 \Rightarrow \gamma = 1$.Intersection point $R_1 = (10-11, 20-21, 30-29) = (-1, -1, 1)$.Thus,$\vec{OR_1} = -\hat{i}-\hat{j}+\hat{k}$.Normal vector $\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 3 & 2 & \gamma \end{vmatrix} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 3 & 2 & 1 \end{vmatrix} = \hat{i}(2-6) - \hat{j}(1-9) + \hat{k}(2-6) = -4\hat{i} + 8\hat{j} - 4\hat{k}$.Unit normal $\hat{n} = \pm \frac{-4\hat{i} + 8\hat{j} - 4\hat{k}}{\sqrt{16+64+16}} = \pm \frac{-4\hat{i} + 8\hat{j} - 4\hat{k}}{\sqrt{96}} = \pm \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}}$.Matching with options,we take $\hat{n} = \frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k}$.$\vec{OR_1} \cdot \hat{n} = (-\hat{i}-\hat{j}+\hat{k}) \cdot (\frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k}) = \frac{-1+2+1}{\sqrt{6}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{4}{6}} = \sqrt{\frac{2}{3}}$.Therefore,$(P) ightarrow (3), (Q) ightarrow (4), (R) ightarrow (1), (S) ightarrow (5)$.

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

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