If $A(2,1,-1)$, $B(6,-3,2)$, and $C(-3,12,4)$ are the vertices of a triangle $ABC$, and the equation of the plane containing the triangle $ABC$ is $53x + by + cz + d = 0$, then find the value of $\frac{d}{b+c}$.

Let the foot of the perpendicular drawn from the point $(1, 2, 3)$ to a plane be $(-1, 3, -2)$. Then the perpendicular distance from the origin to the plane is

If the distance between the plane,$23x - 10y - 2z + 48 = 0$ and the plane containing the lines $\frac{x+1}{2} = \frac{y-3}{4} = \frac{z+1}{3}$ and $\frac{x+3}{2} = \frac{y+2}{6} = \frac{z-1}{\lambda}$ $(\lambda \in R)$ is equal to $\frac{k}{\sqrt{633}}$,then $k$ is equal to

If $(2, -1, 3)$ is the foot of the perpendicular drawn from the origin $(0, 0, 0)$ to a plane,then the equation of that plane is:

$A$ plane $x$ passes through the point $(1, 1, 1)$. If $b, c, a$ are the direction ratios of a normal to the plane,where $a, b, c$ $(a < b < c)$ are the factors of $2001$,then the equation of the plane is

The distance of the point $(1,3,-7)$ from the plane passing through the point $(1,-1,-1)$ having normal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$ is