If from a point $P(a, b, c)$, perpendiculars $PA$ and $PB$ are drawn to $YZ$ and $ZX$ planes respectively, then the equation of the plane $OAB$ is

$A$ plane meets the coordinate axes at points $A$, $B$ and $C$, such that the centroid of triangle $ABC$ is $(2, -\frac{2}{3}, \frac{1}{2})$. The perpendicular distance from the origin to this plane is:

$A$ vector $\vec{n}$ is inclined to the $x$-axis at $45^\circ$,to the $y$-axis at $60^\circ$,and at an acute angle to the $z$-axis. If $\vec{n}$ is a normal to a plane passing through the point $(\sqrt{2}, -1, 1)$,then the equation of the plane is:

$A$ tetrahedron has vertices $O(0,0,0)$, $A(1,2,1)$, $B(2,1,3)$, and $C(-1,1,2)$. If $\theta$ is the angle between the faces $OAB$ and $ABC$, then $\cos \theta =$

The direction ratios of the two lines $AB$ and $AC$ are $1, -1, -1$ and $2, -1, 1$. The direction ratios of the normal to the plane $ABC$ are

The length of the perpendicular from the origin to the plane $\bar{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 14$ is