$\pi$ is a plane passing through the origin and containing two lines whose direction ratios are $1, -2, 2$ and $2, 3, -1$. Then,the direction ratios of the line of intersection of the planes $x - y - z + 1 = 0$ and $\pi$ are:

The line $\frac{x - 1}{2} = -(y + 1) = \frac{z}{3}$ and the plane $3x + 2y - z = 5$ intersect at a point. The coordinates of the point are:

Let $L$ be the line passing through the points $\hat{i}-9 \hat{k}$ and $7 \hat{j}+\hat{k}$ and $\pi$ be the plane passing through the point $6 \hat{i}+\hat{j}$ and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$. If $\theta$ is the angle between $L$ and $\pi$,then $\sin \theta=$

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:

Find the equation of the plane passing through the line of intersection of the planes $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$ and parallel to the $x$-axis.

The angle between the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and the plane $3x + 2y - 3z = 4$ is ......... $^o$