The distance of the origin from the plane $r \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})=7$ measured parallel to the line $r=(\hat{i}+2 \hat{j}+3 \hat{k})+t(6 \hat{i}+2 \hat{j}+3 \hat{k})$ is

Let $\Pi$ be a plane containing the points $(0,-5,-1), (1,-2,5), (-3,5,0)$ and $L$ be a line passing through the point $(0,-5,-1)$ and parallel to the vector $\hat{i}+5\hat{j}-6\hat{k}$. Then the length of the projection of the unit normal vector to the plane $\Pi$ on the line $L$ is

If $P(2, \beta, \alpha)$ lies on the plane $x+2y-z-2=0$ and $Q(\alpha, -1, \beta)$ lies on the plane $2x-y+3z+6=0$,then the direction cosines of the line $PQ$ are:

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

The sine of the angle between the straight line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ and the plane $2x-2y+z=5$ is

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: