Consider the planes $3x - 6y - 2z = 15$ and $2x + y - 2z = 5$. Statement - I : The parametric equations of the line of intersection of the given planes are $x = 3 + 14t$, $y = 1 + 2t$, $z = 15t$. because Statement - II : The vector $14\hat{i} + 2\hat{j} + 15\hat{k}$ is parallel to the line of intersection of given planes.

$\text{Given: } P_1:\ 3x - 6y - 2z = 15, \quad P_2:\ 2x + y - 2z = 5$

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

$\text{Statement-I: line } \vec{r} = (3,1,0) + t(14,2,15)$

$\text{Statement-II: } 14\hat{i}+2\hat{j}+15\hat{k} \parallel L$

$\text{To check: are I and II true, and does II explain I?}$

$\text{Plan: } 1.\ \vec{d} = \vec{n}_1\times\vec{n}_2 \Rightarrow \text{check Statement-II}$ $2.\ (3,1,0) \in P_1,\ P_2\,? \Rightarrow \text{check Statement-I}$ $L \subset P_1,\ L \subset P_2 \Rightarrow \vec{d}\perp\vec{n}_1,\ \vec{d}\perp\vec{n}_2$ $\vec{d} = \vec{n}_1\times\vec{n}_2$ $\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & -2 \\ 2 & 1 & -2 \end{vmatrix}$ $= \hat{i}(12+2) - \hat{j}(-6+4) + \hat{k}(3+12)$ $\vec{d} = 14\hat{i} + 2\hat{j} + 15\hat{k}$ $\vec{n}_1\cdot\vec{d} = 3(14) - 6(2) - 2(15) = 42-12-30 = 0\ \checkmark$ $\vec{n}_2\cdot\vec{d} = 2(14) + 1(2) - 2(15) = 28+2-30 = 0\ \checkmark$ $\vec{d} = 14\hat{i}+2\hat{j}+15\hat{k} \Rightarrow \text{Statement-II is TRUE}$ $\text{At } t=0:\ (x,y,z) = (3,1,0)$ $P_1:\ 3(3) - 6(1) - 2(0) = 3 \neq 15\ \text{✗}$ $P_2:\ 2(3) + 1 - 2(0) = 7 \neq 5\ \text{✗}$ $\text{Statement-I line is PARALLEL to } L \text{, not } L\ \text{✗}$ $P_1 - P_2:\ (3x-2x) + (-6y-y) + (-2z+2z) = 15-5$ $\Rightarrow x - 7y = 10 \Rightarrow x = 10 + 7y$ $y=1 \Rightarrow x = 17;\quad P_2:\ 34 + 1 - 2z = 5$ $\Rightarrow z = 15 \Rightarrow (17,1,15) \in L$ $L:\ \vec{r} = (17,1,15) + t(14,2,15)$ $\text{Statement-I uses the wrong point} \Rightarrow \text{FALSE}$ $\text{Statement-I: False, Statement-II: True} \Rightarrow \text{Option (4)}$