The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is

Let the equation of the plane passing through the line of intersection of the planes $x+2y+az=2$ and $x-y+z=3$ be $5x-11y+bz=6a-1$. For $c \in \mathbb{Z}$,if the distance of this plane from the point $(a, -c, c)$ is $\frac{2}{\sqrt{a}}$,then $\frac{a+b}{c}$ is equal to

$X$ 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

The foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $\pi_1$ is $(1, 3, 5)$. If $(2, 2, -1), (3, 4, 2), (3, 3, 0)$ are three points on the plane $\pi_2$, then the angle between the planes $\pi_1$ and $\pi_2$ is

If the equation of the plane passing through the line of intersection of the planes $2x - 7y + 4z - 3 = 0$ and $3x - 5y + 4z + 11 = 0$ and the point $(-2, 1, 3)$ is $ax + by + cz - 7 = 0$,then the value of $2a + b + c - 7$ is

Let $L$ be the line of intersection of planes $\vec{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=2$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})=2$. If $P(\alpha, \beta, \gamma)$ is the foot of perpendicular on $L$ from the point $(1,2,0)$,then the value of $35(\alpha+\beta+\gamma)$ is equal to :