Let $P\left(x_1, y_1, z_1\right)$ be the foot of perpendicular drawn from the point $Q(2,-2,1)$ to the plane $x-2 y+z=1$. If $d$ is the perpendicular from the point $Q$ to the plane and $l=x_1+y_1+z_1$, then $l+3 d^2$ is

$A(1,2,1), B(2,3,2), C(3,1,3)$ and $D(2,1,3)$ are the vertices of a tetrahedron. If $\theta$ is the angle between the faces $A B C$ and $A B D$, then $\cos \theta=$

Consider the tetrahedron with the vertices $A(3,2,4)$, $B\left(x_1, y_1, 0\right), C\left(x_2, y_2, 0\right)$ and $D\left(x_3, y_3, 0\right)$.If the $\triangle B C D$ is formed by the lines $y=x, x+y=6$ and $y=1$, then the centroid of the tetrahedron is

If $P(2, \beta, \alpha)$ lies on the plane $x+2 y-z-2=0$ and $Q(\alpha,-1, \beta)$ lies on the plane $2 x-y+3 z+6=0$, then the direction cosines of the $P Q$ are