Find the angle between the planes $x + y + 2z = 9$ and $2x - y + z = 15$.

If the plane $\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1$ cuts the coordinate axes at points $A, B$,and $C$,then the area of the triangle $ABC$ is

The Cartesian equation of a plane which passes through the points $A(2,2,2)$ and makes equal non-zero intercepts on the coordinate axes is

The plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ cuts the coordinate axes at the points $A, B, C$ respectively. Then the area of triangle $ABC$ is

$A$ plane $\pi_1$ passing through the point $3 \hat{i}-7 \hat{j}+5 \hat{k}$ is perpendicular to the vector $\hat{i}+2 \hat{j}-2 \hat{k}$ and another plane $\pi_2$ passing through the point $2 \hat{i}+7 \hat{j}-8 \hat{k}$ is perpendicular to the vector $3 \hat{i}+2 \hat{j}+6 \hat{k}$. If $p_1$ and $p_2$ are the perpendicular distances from the origin to the planes $\pi_1$ and $\pi_2$ respectively, then $p_1-p_2=$

Find the distance of the point $(0, 0, 0)$ from the plane $3x - 4y + 12z = 3$. (in $/13$)