Find the equation of the plane whose intercepts on the coordinate axes are $-4, 2$,and $3$.

The equation of the plane passing through the point $(1, 2, -3)$ and perpendicular to the planes $3x + y - 2z = 5$ and $2x - 5y - z = 7$ 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 $a^2 + b^2 + c^2 =$

$A$ plane passing through the points $A(2, 3, 5)$ and $B(-3, -5, -7)$ is perpendicular to the plane $x - y + z = 1$. Which of the following points lies on this plane?

If the foot of the perpendicular drawn from the point $P(2,0,-3)$ to the plane $\pi$ is $F(1,-2,0)$ and the equation of the plane $\pi$ is $ax+by-3z+d=0$,then $a+b+d=$

The direction ratios of the normal to the plane passing through the origin and the line of intersection of the planes $x+2y+3z=4$ and $4x+3y+2z=1$ are . . . . . .