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Q.
The point of intersection of the lines $\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}$ and $\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}$ is
Three Dimensional Geometry
Solution:
Given lines are
$\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}=r_{1}$ (say)
and $\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}=r_{2}$ (say)
$\therefore x =3 r_{1}+5=-36 r_{2}-3$,
$ y =-r_{1}+7=3+2 r_{2} $
and $z =r_{1}-2=4 r_{2}+6$
On solving, we get
$x=21, y=\frac{5}{3}, z=\frac{10}{3}$