How to find this locus? (Analytic Geometry)

Question

I'm stuck with this question, if someone can give some help I would appreciate it!

Find the locus of the midpoints of the line segments with extremities on the planes given by: $P: 2x-3y+3z-4=0$
$P': x-y-z+2=0$

Answer 1

There are not enough conditions to find the locus. Unless the question is changed to from one point on one of the planes say P'.

Let us use the point (1, 1, 2) on P'. Then for$(x_1, y_1, z_1)$ on plane P and $(x, y, z)$ on the locus we have

$$x = \frac{1}{2}(x_1 + 1)$$ $$y = \frac{1}{2}(y_1 + 1)$$ $$z = \frac{1}{2}(z_1 + 2) = \frac{1}{2}(\frac{4 - 2x_1 + 3y_1}{3} + 2)$$ $$= \frac{1}{6}(4 - 2x_1 + 3y_1 + 6)$$ $$= \frac{1}{6}(10 - 2(2x - 1) + 3(2y - 1))$$ $$\implies 4x - 6y + 6z - 9 = 0$$