Find the distance from the origin to the plane passing through the point $(2, 3, -1)$ and perpendicular to the vector $3\hat{i} - 4\hat{j} + 7\hat{k}$.

$A$ plane meets the coordinate axes at $P, Q, R$ respectively. If the centroid of $\triangle P Q R$ is $\left(1, \frac{1}{2}, \frac{1}{3}\right)$, then the equation of the plane is

Find the direction cosines of the unit vector perpendicular to the plane $\vec{r} \cdot (6 \hat{i} - 3 \hat{j} - 2 \hat{k}) + 1 = 0$.

If the foot of the perpendicular drawn from the origin to a plane is $P(-1, -1, 2)$,then the equation of the plane is

The image point of $(1, 3, 4)$ in the plane $2x - y + z + 3 = 0$ 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 $p_1-p_2=$