Find the equation of the plane containing the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ and the point $(0, 7, -7)$.

If the lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar,then $k$ can have:

The value of $k$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$ is

Let $\pi_1$ be the plane determined by the vectors $\hat{i}+\hat{j}$ and $\hat{j}+\hat{k}$,and $\pi_2$ be the plane determined by the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{j}-\hat{k}$. Let $\vec{a}$ be a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If $|\vec{a}|=\sqrt{14}$,then $|\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})|=$

The line $\ell$ passes through the point $P(2, 1, 1)$ and is parallel to the plane $x + y + 2z = 18$. If line $\ell$ intersects the line $L_2: \frac{x + 2}{3} = \frac{y + 1}{-1} = \frac{z - 2}{1}$, then the equation of the line $\ell$ is:

Let the line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the mid-point of the line segment $AB$ from the plane $2x-2y+z=14$ is