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

$A$ variable plane passes through a fixed point $(3, 2, 1)$ and meets the $x, y,$ and $z$ axes at $A, B,$ and $C$ respectively. $A$ plane is drawn parallel to the $yz$-plane through $A$,a second plane is drawn parallel to the $zx$-plane through $B$,and a third plane is drawn parallel to the $xy$-plane through $C$. Then,the locus of the point of intersection of these three planes is:

The equation of the plane passing through $(1, -1, 2)$ and perpendicular to the planes $x + 2y - 2z = 4$ and $3x + 2y + z = 6$ is:

The equations of planes parallel to the plane $x+2y+2z+8=0$,which are at a distance of $2$ units from the point $(1,1,2)$ are

Let $A=(-3,-2,7)$ and $B=(3,1,-2)$. Let a plane perpendicular to the line segment $AB$ divide $AB$ in the ratio $2:1$. Then the intercept made by the plane on the $y$-axis is

Find the vector and Cartesian equations of the plane that passes through the point $(1, 0, -2)$ and the normal to the plane is $\hat{i} + \hat{j} - \hat{k}$.