Show that the co-ordinates of the point on the join of $(-3, 7, -13)$ and $(-6, 1, -10)$ which is nearest to the intersection of the planes $3x-y- 3z + 32 =0$ and $3x+2y-15z= 8$ is $(-7,-1,-9)$.
Same question once again Find the co-ordinates of the point on the join of two points which is nearest to the intersection of two planes I have asked this question before,. but want to solve without using vector method.
Attempt:
I have tried to solve it without using vector method.
The 1st straight line joining $(-3, 7, -13)$ and $(-6, 1, -10)$ in symmetrical form is $$\frac{x+3}{1}=\frac{y-7}{2}=\frac{z+13}{-1}=r_1 (\text{say})$$ Any point on 1st straight line $P(r_1-3, 2r_1+7, -r_1-13)$.
The second straight line in symmetrical form is $$\frac{x+\frac{56}{9}}{7}=\frac{y-\frac{40}{3}}{12}=\frac{z-0}{3}=r_2 (\text{say})$$
Any point on 2nd straight line $P(7r_2-\frac{56}{9}, 12r_2+\frac{40}{3}, 3r_2)$.
As P is nearest to the intersection of the planes, consider PQ is the shortest distance(SD) from P on the 2nd straight line. By the definition of SD, PQ is perpendicular to both the lines written in symmetrical form. The direction cosine of th PQ are $$r_{{1}}-7\,r_{{2}}+{\frac {29}{9}},~~2\,r_{{1}}-12\,r_{{2}}-19/3,~~ -r_{{1}}-3\,r_{{2}}-13$$
As PQ is perpendicular to both the lines, then using the condition of perpendicularity, we have $$3\,r_{{1}}-14\,r_{{2}}+{\frac {16}{9}}=0$$ and $$17\,r_{{1}}-92\,r_{{2}}-{\frac {65}{9}}=0$$. Solving $$r_1 = -397/57, r_2 = -467/342$$
Then the coordinate of $P$ is not same than that is asked.
This answer is not same what is expected. Please help me to answer the problem wihout using verctor metheod rather using 3D geometry laws.