Find the coordinates of the foot of the perpendicular drawn from the point $P(-1, 1, 2)$ to the plane $2x - 3y + z - 11 = 0$.

The line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the curve $xy = c^2, z = 0$ if $c$ is equal to

Let $d$ be the distance between the foot of perpendiculars of the points $P(1, 2, -1)$ and $Q(2, -1, 3)$ on the plane $-x + y + z = 1$. Then $d^{2}$ is equal to

Three lines are given by $\overrightarrow{r} = \lambda \hat{i}, \lambda \in R$,$\overrightarrow{r} = \mu(\hat{i} + \hat{j}), \mu \in R$ and $\overrightarrow{r} = v(\hat{i} + \hat{j} + \hat{k}), v \in R$. Let the lines cut the plane $x + y + z = 1$ at the points $A, B$ and $C$ respectively. If the area of the triangle $ABC$ is $\Delta$,then the value of $(6 \Delta)^2$ equals.

Find the equation of the plane passing through the intersection of the planes $P_1$ and $P_2$ and parallel to the line $L$,where:
$P_1 : 3x + 2y + 5z + 1 = 0$
$P_2 : x + y + z + 2 = 0$
$L : \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3y-\alpha z+\beta=0$. Then the value of $(\beta-\alpha)$ is equal to