The distance of a point $(2, 3, -5)$ from the plane $\vec{r} \cdot (4 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 4$ is

The equation of the plane passing through $(-1, 1, 2)$ whose normal makes equal acute angles with the coordinate axes is:

The equation of the planes parallel to the plane $x - 2y + 2z - 3 = 0$ which are at a unit distance from the point $(1, 2, 3)$ is $ax + by + cz + d = 0$. If $(b - d) = K(c - a)$,then the positive value of $K$ is

$A$ plane $\pi$ makes intercepts $3$ and $4$ respectively on $Z$-axis and $X$-axis. If $\pi$ is parallel to $Y$-axis, then its equation is:

The equation of the plane passing through the points with position vectors $A(2 \hat{i}+6 \hat{j}-6 \hat{k})$, $B(-3 \hat{i}+10 \hat{j}-9 \hat{k})$ and $C(-5 \hat{i}-6 \hat{k})$ is

Direction ratios of the normal to the plane passing through the points $A(0, 1, 1)$,$B(1, 1, 2)$,and $C(-1, 2, -2)$ are