The line given by the equations $x-2y+4z+4=0$ and $x+y+z-8=0$ intersects the plane $x-y+2z+1=0$ at the point:

The line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is parallel to the plane

If the line $\frac{x+1}{1}=\frac{y-k}{11}=\frac{z-4}{-5}$ lies in the plane $2x+py+7z-41=0$ which is perpendicular to the plane $x+4y-2z+13=0$,then $k=$

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

The equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $3x+4y+5z=2$ and perpendicular to the $XY$-plane is

$A$ line $L$ is parallel to both the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If line $L$ makes an angle $\alpha$ with the positive direction of the $X$-axis, then $\cos \alpha =$