# Finding the intersection of a line with $x,y,z$ coordinate planes

How do I find the points at which the line given by the symmetric equation: $$\frac{-x - 1}{-1} = \frac{y + 5}{2} = \frac{z - 6}{-3}$$ intersects the coordinate planes $xz$, $yz$, $xy$?

-

One idea: First write the line as parametric equations: \begin{align} x &= -1 +t \\ y &= -5 + 2t \\ z &= 6 - 3t. \end{align} You have intersection with $xy$-plane when $z=0$, so that gives $t=2$ which gives you the point of intersection $(1, -1, 0)$. Now do likewise for the other planes.
Hint: Just do it. For example, the intersection with the coordinate plane $xz$ means that $y=0$, which gives
$$\frac {-x-1}{-1} = \frac {0 + 5}{2} = \frac {z-6}{-3}.$$