The distance of the point $(-1, -5, -10)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x-y+z=5$ is

What is the distance of the point $(2, 3, 4)$ from the plane $3x - 6y + 2z + 11 = 0$?

The equation of a plane containing the lines $\overline{r}=(\hat{\imath}+2 \hat{\jmath}-4 \hat{k})+\lambda(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k})$ and $\overline{r}=(\hat{\imath}+3 \hat{\jmath}+4 \hat{k})+\mu(\hat{\imath}+\hat{\jmath}-\hat{k})$ is

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:

Show that the lines $\frac{x-a+d}{\alpha-\delta}=\frac{y-a}{\alpha}=\frac{z-a-d}{\alpha+\delta}$ and $\frac{x-b+c}{\beta-\gamma}=\frac{y-b}{\beta}=\frac{z-b-c}{\beta+\gamma}$ are coplanar.

If the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}$ and $\frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1}$ intersect at the point $P$,then the distance of the point $P$ from the plane $z = a$ is: