The equation of a plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is

$A$ plane ( $\pi$ ) passing through the point $(1, 2, -3)$ is perpendicular to the planes $x + y - z + 4 = 0$ and $2x - y + z + 1 = 0$. If the equation of the plane ( $\pi$ ) is $ax + by + cz + 1 = 0$, then $a^2 + b^2 + c^2 =$

The plane $ax + by + cz = 1$ meets the coordinate axes at points $A, B$,and $C$. Find the centroid of the triangle $ABC$.

If planes $x - c y - b z = 0$,$c x - y + a z = 0$ and $b x + a y - z = 0$ pass through a straight line then $a^2 + b^2 + c^2 =$

The vector equation of the plane $r = (2 \hat{i} + \hat{k}) + \lambda(\hat{i}) + \mu(\hat{i} + 2 \hat{j} - 3 \hat{k})$ in scalar product form is $r \cdot (3 \hat{i} + 2 \hat{k}) = \alpha$,then $\alpha = \dots$

$A$ line joining the points $(1, 2, 0)$ and $(4, 13, 5)$ is perpendicular to a plane. Then the coefficients of $x, y$ and $z$ in the equation of the plane are respectively