The sum of intercepts of the plane $4x + 3y + 2z = 2$ on the coordinate axes is

$A$ plane $\pi$ passes through the points $(5,1,2)$, $(3,-4,6)$, and $(7,0,-1)$. If $p$ is the perpendicular distance from the origin to the plane $\pi$ and $l, m, n$ are the direction cosines of a normal to the plane $\pi$, then $|3l+2m+5n|=$

If the plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{3} = 1$ intersects the coordinate axes at $A, B, C$,then the area of $\Delta ABC$ is:

The vector equation of the plane $r = (2 \hat{i} + \hat{k}) + \lambda(\hat{i}) + \mu(\hat{i} + 2 \hat{j} - 3 \hat{k})$ in scalar product form is $r \cdot (3 \hat{i} + 2 \hat{k}) = \alpha$,then $\alpha = \dots$

$A$ plane is parallel to two lines,whose direction ratios are $1, 0, -1$ and $-1, 1, 0$ and it contains the point $(1, 1, 1)$. If it cuts coordinate axes ($X, Y, Z$-axes respectively) at $A, B, C$,then the volume of the tetrahedron $OABC$ is (in cubic units):

The coordinates of the foot of the perpendicular drawn from the origin to the plane $3x + 2y + 6z = 56$ are: