Find the equation of the plane passing through the origin and parallel to the plane $3x - 4y + 5z - 6 = 0$.

The equation of a plane parallel to the $x$-axis is:

If the equation of the plane which is at a distance of $\frac{1}{3}$ units from the origin and perpendicular to a line whose directional ratios are $(1, 2, 2)$ is $x+py+qz+r=0$,then $\sqrt{p^2+q^2+r^2}=$

The vector equation of the plane containing the point $(1, -1, 2)$ and perpendicular to the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$ is:

The plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ cuts the coordinate axes at the points $A, B, C$ respectively. Then the area of triangle $ABC$ is

If $O$ is the origin and $A$ is the point $(a, b, c)$,then the equation of the plane passing through $A$ and perpendicular to $OA$ is: