The equation of a plane which passes through the point of intersection of lines $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}, and \frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}$  and at greatest distance from point (0, 0, 0) is

Let a point $(3\lambda +1, \lambda +2, 2\lambda +3)$ of the first line also lies on the second line. Then
$\frac{3\lambda +1-3}{1}=\frac{\lambda +2-1}{2}=\frac{2\lambda +3-2}{3} or \lambda =1$
Hence, the point of intersection P of the two lines is (4, 3, 5).
Equation of plane perpendicular to OP, where O is (0, 0, 0), 

and passing through P is 4x + 3y + 5z = 50.