Condition of the problem

Find the point of intersection of the line and the plane.
$\frac{x+3}{1}=\frac{y-2}{-5}=\frac{z+2}{3}$

$5 x-y+4 z+3=0$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x+3}{1}=\frac{y-2}{-5}=\frac{z+2}{3}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-3+t \\
y=2-5 t \\
z=-2+3 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$5(-3+t)-(2-5 t)+4(-2+3 t)+3=0$

$-15+5 t-2+5 t-8+12 t+3=0$

$-22+22 t=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=-3+1=-2 \\
y=2-5 \cdot 1=-3 \\
z=-2+3 \cdot 1=1
\end{array}\right.
$$

We get:

$(-2, -3, 1)$

## Problem Kuznetsov Analytic Geometry 14-14