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

Let $O(\overrightarrow{0}), A(\hat{i}+2 \hat{j}+\hat{k}), B(-2 \hat{i}+3 \hat{k}), C(-2 \hat{i}+\hat{j}), D(4 \hat{k})$ be the position vectors of the points $O, A, B, C$ and $D$. If a line passing through $A$ and $B$ intersects the plane passing through $O, C$ and $D$ at the point $R$,then the position vector of $R$ is:

The line of intersection of the planes $x + 2y = 0$ and $y - 3z + 3 = 0$ is

$X$ 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 point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane $2x + 3y + z = 0$ is

The angle made by the vector $2\hat{i}-\hat{j}+\hat{k}$ with the plane represented by $\vec{r} \cdot(\hat{i}+\hat{j}+2\hat{k})=7$ is (in $^{\circ}$)