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

$A$ plane meets the $X, Y, Z$-axes in $A, B, C$ respectively. If the centroid of the $\triangle ABC$ is $(2, -3, 5)$,then the perpendicular distance from the origin to the given plane is:

The equation of a plane containing the point $(1, -1, 2)$ and perpendicular to the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$ is

If the plane $-4x - 2y + 2z + \alpha = 0$ is at a distance of $2$ units from the plane $2x + y - z + 1 = 0$, then the product of all the possible values of $\alpha$ is

$A$ variable plane passes through a fixed point $(\alpha, \beta, \gamma)$ and meets the coordinate axes in $A, B$ and $C$. Let $P_1, P_2$ and $P_3$ be the planes passing through $A, B, C$ and parallel to the coordinate planes $YZ, ZX, XY$ respectively. Then,the locus of the point of intersection of the planes $P_1, P_2$ and $P_3$ is

Let $(\alpha, \beta, \gamma)$ be the image of the point $P (2, 3, 5)$ in the plane $2x + y - 3z = 6$. Then $\alpha + \beta + \gamma$ is equal to