The distance from the point $(-1, -5, -10)$ to 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:

Let the points $P, Q$ and $R$ have position vectors $\overrightarrow{r_1} = 3i - 2j - k, \overrightarrow{r_2} = i + 3j + 4k$ and $\overrightarrow{r_3} = 2i + j - 2k$ respectively relative to an origin $O$. Then the distance of $P$ from the plane $OQR$ is:

If the line passing through the points $(a, 2, -4)$ and $(5, 3, b)$ crosses the $ZX$-plane at the point $(-a+2b, 0, a+b)$, then find the value of $14a+7b$.

Find the angle between the line $\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(\hat{i} + \hat{j} - \hat{k})$ and the plane $\vec{r} \cdot (-2\hat{i} + \hat{j} - \hat{k}) = 0$.

Find the equation of the plane which is perpendicular to the plane $5x + 3y + 6z + 8 = 0$ and which contains the line of intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$.

$A$ vector parallel to the line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is