The square of the distance of the point of intersection of the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}$ and the plane $2x-y+z=6$ from the point $(-1,-1,2)$ is .... .

The angle between the line $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}$ and the plane $\bar{r} \cdot(6 \hat{\imath}-2 \hat{\jmath}-3 \hat{k})=5$ is

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:

The line $\frac{x - 3}{2} = \frac{y - 4}{3} = \frac{z - 5}{4}$ lies in the plane $4x + 4y - kz - d = 0$. The values of $k$ and $d$ are

For real numbers $\alpha$ and $\beta \neq 0$,if the point of intersection of the straight lines $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}$ lies on the plane $x+2y-z=8$,then $\alpha-\beta$ is equal to:

The distance between the line $\vec{r} = (2\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$ is: