Find the distance of the plane $2x - 3y + 4z - 6 = 0$ from the origin.

What are the direction cosines of the normal to the plane $x + 2y - 3z + 4 = 0$?

If $\alpha$ and $\beta$ are scalars and $\vec{r} = (2+\alpha-3\beta) \hat{i} + (\beta-3) \hat{j} + (2\alpha-5\beta-1) \hat{k}$ is the equation of a plane, then its equation in Cartesian form is:

$A$ plane $\pi$ given by $ax + by + 11z + d = 0$ is perpendicular to the planes $2x - 3y + z = 4$ and $3x + y - z = 5$. The perpendicular distance from the origin to the plane $\pi$ is $\sqrt{6}$ units. If all the intercepts made by the plane $\pi$ on the coordinate axes are positive,then $d =$

$\pi_1$ is a plane passing through the point $(1, 2, 3)$ and perpendicular to the planes $x+2y+3z-6=0$ and $x+2y+2z-5=0$. If $(-1, 2, -3)$ is the foot of the perpendicular drawn from the point $(1, 3, 2)$ onto a plane $\pi_2$, then the angle between the planes $\pi_1$ and $\pi_2$ is

The vector equation of the plane passing through the point $A(1, 2, -1)$ and parallel to the vectors $2 \hat{i} + \hat{j} - \hat{k}$ and $\hat{i} - \hat{j} + 3 \hat{k}$ is