The distance of the point $P(1, 0, -3)$ from the plane $x - y - z = 9$ measured parallel to the line $\frac{x - 2}{2} = \frac{y + 2}{3} = \frac{z - 6}{-6}$ is (in $\text{ units}$)

Let $A$ be a point having position vector $\bar{i}-3 \bar{j}$ and $\bar{r}=(\bar{i}-3 \bar{j})+t(\bar{j}-2 \bar{k})$ be a line. If $P$ is a point on this line and is at a minimum distance from the plane $\bar{r} \cdot(2 \bar{i}+3 \bar{j}+5 \bar{k})=0$, then the equation of the plane through $P$ and perpendicular to $AP$ 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:

The line of intersection of the planes $r \cdot (i - 3j + k) = 1$ and $r \cdot (2i + 5j - 3k) = 2$ is parallel to which vector?

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

The equation of the plane passing through the intersection of the planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ and parallel to the line $y = 0, z = 0$ is: