$X$ intercept of the plane containing the line of intersection of the planes $x-2y+z+2=0$ and $3x-y-z+1=0$ and also passing through $(1,1,1)$ is

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

$A$ plane $P$ contains the line of intersection of the planes $\vec{r} \cdot (\hat{i}+\hat{j}+\hat{k}) = 6$ and $\vec{r} \cdot (2\hat{i}+3\hat{j}+4\hat{k}) = -5$. If $P$ passes through the point $(0, 2, -2)$,then the square of the distance of the point $(12, 12, 18)$ from the plane $P$ is

If the point $(1, \alpha, \beta)$ lies on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}, z=1$,then $\alpha+\beta=$

Let the plane containing the line of intersection of the planes $P_1: x+(\lambda+4)y+z=1$ and $P_2: 2x+y+z=2$ pass through the points $(0,1,0)$ and $(1,0,1)$. Then the distance of the point $(2\lambda, \lambda, -\lambda)$ from the plane $P_2$ is (in $\sqrt{6}$)

The equation of the plane through the intersection of the planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to the $x$-axis is: