let there be $4$ points. $A(-1,1,1), B(2,0,-1), C(1,3,-2), D(-2,-1,0)$. the $4$ points are not on the same line. the plane which goes through the points $A$ and $B$, and which is also paralel to the line $CD$, is, from calculations $ 2x+3z-1 = 0$ . The task is to find the symmetric point to D in accordance to the said plane.
I tried this: formula of the line that goes through the two points and is vertical to the plane is $ (-2,-1,0) + t(2,0,3)$ hence a general point would be $(2t-2,-1,3t)$. puting this in the plane formula yields $t=5/13$. this in the line formula yields the middle point between the symmetric points whihc is also on the plane $(-16/13 , -1, 15/13)$. each coordinate is the sum of the aprrprite two coordinates (of same axis) divided by two and from that system I get $(-6/13, -1,30/13)$ which is the second symmetric point and what was required...
offical answer is however $(18, -1 ,-10) $
What am I doing wrong?