The equation of a plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by

If the foot of the perpendicular drawn from the origin to a plane is $M(-1, -2, 2)$,then the vector equation of the plane is

$A$ plane meets the coordinate axes at $P, Q, R$ respectively. If the centroid of $\triangle P Q R$ is $\left(1, \frac{1}{2}, \frac{1}{3}\right)$, then the equation of the plane is

The distance between two parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ is given by $\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$. If the plane $2x-y+2z+3=0$ is at distances of $\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4x-2y+4z+\lambda=0$ and $2x-y+2z+\mu=0$ respectively,then the maximum value of $\lambda+\mu$ is:

$A$ plane passes through the point $A(2, 1, -3)$. If the distance of this plane from the origin is maximum,then its equation is

If the foot of the perpendicular from $(0,0,0)$ to the plane is $(1,2,2)$, then the equation of the plane is