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):

If a plane $x+y+z-5=0$ intersects the line joining $A(1,1,1)$ and $B(2,2,2)$ at $P$, then $AP: PB=$

Let a line $L$ passing through the point $P(1, 1, 1)$ be perpendicular to the lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line $L$ intersect the $yz$-plane at the point $Q$. Another line parallel to $L$ and passing through the point $S(1, 0, -1)$ intersects the $yz$-plane at the point $R$. Then the square of the area of the parallelogram $PQRS$ is equal to . . . . . . .

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 plane $2x - 3y + 6z - 11 = 0$ makes an angle $\sin^{-1}(\alpha)$ with the $X$-axis. The value of $\alpha$ is equal to:

The point of intersection of the line passing through the points $\hat{i}-\hat{j}$ and $\hat{j}-\hat{k}$ and the plane passing through the points $2 \hat{i}+\hat{j}$, $2 \hat{j}-\hat{k}$, and $\hat{i}+2 \hat{k}$ is