Consider the lines $L_1 : \frac{x-1}{2} = \frac{y}{-1} = \frac{z+3}{1},\ L_2 : \frac{x-4}{1} = \frac{y+3}{1} = \frac{z+3}{2}$ and the planes $P_1 : 7x + y + 2z = 3,\ P_2 : 3x + 5y - 6z = 4$. Let $ax + by + cz = d$ be the equation of the plane passing through the point of intersection of lines $L_1$ and $L_2$, and perpendicular to planes $P_1$ and $P_2$. Match List I with List II and select the correct answer using the code given below the lists : | List I | List II | | --- | --- | | (P) $a =$ | (1) $13$ | | (Q) $b =$ | (2) $-3$ | | (R) $c =$ | (3) $1$ | | (S) $d =$ | (4) $-2$ |

Consider the lines L₁ : (x-1)/2 = y/(-1) = (z+3)/1, L₂ : (x-4)/1 = (y+3)/1 = (z+3)/2 and the planes P₁ : 7x + y + 2z = 3, P₂ : 3x + 5y - 6z = 4. Let ax + by + cz = d be the equation of the plane passing through the point of intersection of lines L₁ and L₂, and perpendicular to planes P₁ and P₂. Match List I with List II and select the correct answer using the code given below the lists : | List I | List II | | --- | --- | | (P) a = | (1) 13 | | (Q) b = | (2)

$L_1 : \frac{x-1}{2} = \frac{y}{-1} = \frac{z+3}{1}, \quad L_2 : \frac{x-4}{1} = \frac{y+3}{1} = \frac{z+3}{2}$

$P_1 : 7x+y+2z=3, \quad P_2 : 3x+5y-6z=4$

$\text{To find: plane } ax+by+cz=d \perp P_1,\ P_2 \text{ through } L_1 \cap L_2$

$\text{Match } a,\ b,\ c,\ d \text{ with List II}$

$L_1: A(1,0,-3), \quad \vec{d}_1 = (2,-1,1)$

$L_2: B(4,-3,-3), \quad \vec{d}_2 = (1,1,2)$

$L_1: (1+2s,\ -s,\ -3+s)$ $L_2: (4+t,\ -3+t,\ -3+2t)$

$2s - t = 3, \quad s + t = 3, \quad s = 2t$

$s = 2, \qquad t = 1$

$L_1(2) = (1+4,\ -2,\ -3+2) = (5,-2,-1)$

$L_2(1) = (4+1,\ -3+1,\ -3+2) = (5,-2,-1)$

$\vec{n}_1 = (7,1,2), \qquad \vec{n}_2 = (3,5,-6)$

$\vec{n} = \vec{n}_1 \times \vec{n}_2$

n⃗ = î ĵ k̂ 7 1 2 3 5 -6

$= (-6-10)\hat{i} - (-42-6)\hat{j} + (35-3)\hat{k}$

$\vec{n} = -16\hat{i} + 48\hat{j} + 32\hat{k}$

$\vec{n} \propto (-16,48,32) \propto (-1,3,2) \propto (1,-3,-2)$

$1(x-5) - 3(y+2) - 2(z+1) = 0$

$x - 3y - 2z = 13$

$ax+by+cz=d \equiv x-3y-2z=13$

$a=1, \quad b=-3, \quad c=-2, \quad d=13$

$a=1 \to (3), \quad b=-3 \to (2)$

$c=-2 \to (4), \quad d=13 \to (1)$

$(P)\to(3),\ (Q)\to(2),\ (R)\to(4),\ (S)\to(1) \ \Rightarrow \ \text{option (1)}$