The direction ratios of the line of intersection of the planes $x-y+z-5=0$ and $x-3y-6=0$ are:

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

Let $\pi_1$ be the plane determined by the vectors $\bar{i}+\bar{j}$ and $\bar{i}+\bar{k}$, and $\pi_2$ be the plane determined by the vectors $\bar{j}-\bar{k}$ and $\bar{k}-\bar{i}$. Let $\bar{a}$ be a non-zero vector parallel to the line of intersection of the planes $\pi_1$ and $\pi_2$. If $\bar{b}=\bar{i}+\bar{j}-\bar{k}$, then the angle between the vectors $\bar{a}$ and $\bar{b}$ is:

Let $P(3, 2, 6)$ be a point in space and $Q$ be a point on the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$. Then the value of $\mu$ for which the vector $\vec{PQ}$ is parallel to the plane $x - 4y + 3z = 1$ is

If the plane passing through the points $\hat{i}+\hat{j}+\hat{k}$, $2\hat{i}-\hat{k}$ and the origin meets the line passing through the points $\hat{i}+3\hat{j}-2\hat{k}$ and $\hat{i}-\hat{j}+3\hat{k}$ at the point $A$, then $A=$

The distance of the point $(3, 8, 2)$ from the line $\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 2}{3}$ measured parallel to the plane $3x + 2y - 2z = 0$ is