The coordinates of the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane $2x - 3y - 6z = 49$ are

Find the coordinates of the point where the line passing through $(3, -4, -5)$ and $(2, -3, 1)$ intersects the plane $2x + y + z = 7$.

The coordinates of the point where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}$ meets the plane $2x+4y-z=1$ are

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if

The distance of the point $(1, 6, 2)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x-y+z=16$ is (in $\text{ units}$)

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 :