In general, given three vectors, there does not exist a plane containing those three vectors. So I suspect you P, Q, R are actually three points, or the position vectors of those three points, and you want to find the plane containing those three points. If your points are given by (equivalently vector ), ( ), and ( ) then vectors in the plane are given by the difference of two of those. In particular, and are vectors in the plane. Their cross product will give a vector, , perpendicular to the plane. And the equation of a plane having that normal vector, containing point is given by .
The shortest distance from a point to a plane (or line) is along a perpendicular line. Once you have the plane as , you know that is perpendicular to the plane and so x= At, y= Bt, z= Ct is a line perpendicular to the plane, through (0, 0, 0). Replace x, y, and z in the equation of the plane and you have a single equation to solve for t, giving the point where that line hits the plane. The distance between that point and (0, 0, 0) is the shortest distance from (0, 0, 0) to the plane.
(If you are doing this for a course, I bet all of that is in your textbook!)