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.

$\text{Given: } \pi_1: 3x-6y-2z=15,\qquad \pi_2: 2x+y-2z=5$ $\text{S-I: } x=3+14t,\quad y=1+2t,\quad z=15t$ $\text{S-II: } 14\hat{i}+2\hat{j}+15\hat{k} \parallel \text{line of intersection}$ $\text{To find: which option about S-I and S-II is correct}$ $\text{Plan: direction} = \vec n_1\times\vec n_2 \Rightarrow \text{test S-II};\ \text{point on line} \Rightarrow \text{test S-I}$ $\pi_1 \Rightarrow \vec n_1 = (3,-6,-2)$ $\pi_2 \Rightarrow \vec n_2 = (2,1,-2)$ $\vec n_1\times\vec n_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & -2 \\ 2 & 1 & -2 \end{vmatrix} = (14,\ 2,\ 15)$ $\therefore\ 14\hat{i}+2\hat{j}+15\hat{k} \parallel \text{line} \Rightarrow \text{S-II is TRUE}$ $\text{Direction of the line of intersection} = (14,2,15)$ $\text{Set } z=0:\quad 3x-6y=15,\quad 2x+y=5$ $\Rightarrow x=3,\quad y=-1 \Rightarrow (3,-1,0)$ $\text{True line: } x=3+14t,\quad y=-1+2t,\quad z=15t$ $\text{S-I uses } (3,1,0):\ 3(3)-6(1)-2(0)=3 \ne 15$ $\therefore\ \text{S-I is FALSE}$ $\text{S-I FALSE},\ \text{S-II TRUE} \Rightarrow \text{option }(4)$ $\boxed{(4)}$