Consider the planes 3x - 6y - 2z = 15 and 2x + y -2z = 5. Statement-1…

The vectors normals to the given planes are $\vec{n}_1 3\hat{i} - 6\hat{j} - 2\hat{k} $ and $\vec{n}_2= 2\hat{i} + \hat{j} - 2\hat{k}$ respectively.

So, a vector parallel to the line of intersection of the given planes is

$\vec{b} = \vec{n}_1 × \vec{n}_2 =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\3 & -6 & -2\\2 & 1 & -2\end{vmatrix}= 14\hat{i} + 2\hat{j} + 15 \hat{k}$

So, statement-2 is true.

The coordinates of the point where the line of intersection of the given planes intersect with xy-plane i.e. z=0 are given by 

$3x -6y = 15 $ and $2x + y = 5 $

Solving these two equations, we get x = 3, y = -1

So, the parametric equations of the line of intersection of the given planes is given by $\frac{x-3}{14}=\frac{y+1}{2}=\frac{z-0}{15}= t, t \, \in \, R.$

or, $ x = 3 + 14 t, y = 2t - 1, z = 15t, t \, \in , R$

So statement-1 is not true.