The distance of the point $(7,5,2)$ from the plane $3x+4y+z-8=0$ measured parallel to the line $\frac{x-1}{3}=\frac{y-2}{6}=\frac{z+1}{2}$ is:

If the line $\frac{x - 1}{2} = \frac{y + \alpha}{\alpha} = \frac{z - \beta}{2}$ lies in the plane $2x + y + z = 5$,then $\alpha + \beta$ is

The equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

The distance of the point having position vector $\hat{i}-2 \hat{j}-6 \hat{k}$ from the straight line passing through the point $(2, -3, -4)$ and parallel to the vector $6 \hat{i}+3 \hat{j}-4 \hat{k}$ is units.

The equation of the plane passing through the points $(3, 2, 2)$ and $(1, 0, -1)$ and parallel to the line $\frac{x - 1}{2} = \frac{y - 1}{-2} = \frac{z - 2}{3}$ 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?