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

The coordinates of the point,where the line through $A(3, 4, 1)$ and $B(5, 1, 6)$ crosses the $XZ$-plane,are

If the distance of the point $P(1, -2, 1)$ from the plane $x + 2y - 2z = \alpha$, where $\alpha > 0$, is $5$, then the foot of the perpendicular from $P$ to the plane is

If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar,then the plane$(s)$ containing these two lines is(are):

Let $\bar{A}$ be a vector parallel to the line of intersection of planes $P_1$ and $P_2$ passing through the origin. $P_1$ is parallel to the vectors $2 \hat{j}+3 \hat{k}$ and $4 \hat{j}-3 \hat{k}$,and $P_2$ is parallel to $\hat{j}-\hat{k}$ and $3 \hat{i}+3 \hat{j}$. Then the angle between $\bar{A}$ and $2 \hat{i}+\hat{j}-2 \hat{k}$ is

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 4$ and $2x + y - z = -5$,and perpendicular to the plane $5x + 3y + 6z + 8 = 0$.