Consider the lines L_1: frac{x-1}{2}=frac{y}{ | vedclass

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(A) The normal vector $\vec{n}$ of the required plane is perpendicular to the normals of $P_1$ and $P_2$. Thus,$\vec{n} = \vec{n_1} 	imes \vec{n_2} =

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$3 \quad 2 \quad 4 \quad 1$

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(A) The normal vector $\vec{n}$ of the required plane is perpendicular to the normals of $P_1$ and $P_2$. Thus,$\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 7 & 1 & 2 \ 3 & 5 & -6 \end{vmatrix} = \hat{i}(-6-10) - \hat{j}(-42-6) + \hat{k}(35-3) = -16\hat{i} + 48\hat{j} + 32\hat{k}$.Dividing by $-16$,we get the normal vector $\vec{n} = \hat{i} - 3\hat{j} - 2\hat{k}$.To find the intersection point of $L_1$ and $L_2$,let $L_1: (2k_1+1, -k_1, k_1-3)$ and $L_2: (k_2+4, k_2-3, 2k_2-3)$.Equating coordinates: $2k_1+1 = k_2+4 \Rightarrow 2k_1 - k_2 = 3$ and $-k_1 = k_2-3 \Rightarrow k_1+k_2 = 3$.Adding these gives $3k_1 = 6 \Rightarrow k_1 = 2$. Then $k_2 = 1$.Point of intersection: $(2(2)+1, -2, 2-3) = (5, -2, -1)$.The plane equation is $1(x-5) - 3(y+2) - 2(z+1) = 0 \Rightarrow x - 3y - 2z - 5 - 6 - 2 = 0 \Rightarrow x - 3y - 2z = 13$.Comparing with $ax+by+cz=d$,we get $a=1, b=-3, c=-2, d=13$.Thus,$P-3, Q-2, R-4, S-1$. The correct option is $A$.

List-$I$	List-$II$
$P. \quad a =$	$1. \quad 13$
$Q. \quad b =$	$2. \quad -3$
$R. \quad c =$	$3. \quad 1$
$S. \quad d =$	$4. \quad -2$

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