## problem statement

Find the point of intersection of the line and the plane.

$\frac{x+2}{1}=\frac{y-2}{0}=\frac{z+3}{0}$
$2 x-3 y-5 z-7=0$
## Solution

Let's write the parametric equations of the line.

$\frac{x+2}{1}=\frac{y-2}{0}=\frac{z+3}{0}=t \Rightarrow$

$\left\{\begin{array}{l}x=-2+t \\ y=2 \\ z=-3\end{array}\right.$

Substitute into the equation of the plane:

$2(-2+t)-3 \cdot 2-5 \cdot(-3)-7=0$

$-4+2 t-6+15-7=0$

$2 t-2=0$

$t=1$

Find the coordinates of the intersection point of the line and the plane:

$$
\left\{\begin{array}{l}
x=-2+1=-1 \\
y=2 \\
z=-3
\end{array}\right.
$$

We get:

$(-1 ; 2 ;-3)$

Problem Kuznetsov Analytic Geometry 14-10