Question

Find the coordinates of the point where the line through the points $(3,-4,-5)$ and $(2,-3,1)$ crosses the plane $3x+2y+z+14=0.$

Answer 1

The line passes through the points $(3,-4,-5)$ and $(2,-3,1)$

The equation of the line can be written as $\frac{x-2}{3-2}=\frac{y+3}{-4+3}=\frac{z-1}{-5-1}=t$

Thus, a point on the line can be written as $(t+2,-t-3,-6t+1)$

This point lies on the plane and so,

$3(t+2)+2(-t-3)+-6t+1+14=0$

$\therefore -5t+15=0$

$\therefore t=3$

Re-substituting t, we get the point as $(5,-3-3,-18+1)=(5,-6,-17)$