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 = | (

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

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

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

$\text{Match } (P)\ a,\ (Q)\ b,\ (R)\ c,\ (S)\ d$

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

$A(1,0,-3),\quad \vec{d_1}=(2,-1,1)$

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

$B(4,-3,-3),\quad \vec{d_2}=(1,1,2)$

$1.\ \text{find } I = L_1 \cap L_2$

$2.\ \vec n = \vec n_1 \times \vec n_2$

$3.\ \text{plane through } I \text{ with normal } \vec n$

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

$\Rightarrow\ 1+2s=4+t,\quad -s=-3+t,\quad -3+s=-3+2t$

$s=2t,\quad t=3-s \Rightarrow s=2(3-s) \Rightarrow s=2,\ t=1$

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

$\vec{n_1}=(7,1,2)\ (\text{from } P_1),\quad \vec{n_2}=(3,5,-6)\ (\text{from } P_2)$

$\vec n \parallel \vec{n_1}\times\vec{n_2}$

n₁⃗×n₂⃗ = î ĵ k̂ 7 1 2 3 5 -6

$= (-16,\ 48,\ 32) = 16(-1,3,2)$

$\vec n=(1,-3,-2):\ (x-5)-3(y+2)-2(z+1)=0$

$x-3y-2z=13\ \Rightarrow\ a=1,\ b=-3,\ c=-2,\ d=13$

$a=1\to(3),\quad b=-3\to(2),\quad c=-2\to(4),\quad d=13\to(1)$

$\text{so } (P)(Q)(R)(S) = 3\ 2\ 4\ 1$

$\boxed{\text{Answer: option (1)} \quad 3\ 2\ 4\ 1}$