If the line $\bar{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})$ is parallel to the plane $\bar{r} \cdot(3 \hat{i}-2 \hat{j}-m \hat{k})=5$,then the value of $m$ is:

The value of $m$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z+m}{2}$ lies in the plane $2x-4y+z=7$ is

Let $P$ be the image of the point $(3, 1, 7)$ with respect to the plane $x-y+z=3$. Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0,7,-7)$ is

Let $P$ be the plane,which contains the line of intersection of the planes $x + y + z - 6 = 0$ and $2x + 3y + z + 5 = 0$ and it is perpendicular to the $xy$-plane. Then the distance of the point $(0, 0, 256)$ from $P$ is equal to

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