## Problem Statement

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

$\frac{x-1}{2}=\frac{y+2}{-5}=\frac{z-3}{-2}$

$x+2 y-5 z+16=0$
## Solution

Let's write the parametric equations of the line.

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

Substitute into the equation of the plane:

$(1+2 t)+2(-2-5 t)-5(3-2 t)+16=0$ $1+2 t-4-10 t-15+10 t+16=0$

$2 t-2=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=1+2 \cdot 1=3 \\
y=-2-5 \cdot 1=-7 \\
z=3-2 \cdot 1=1
\end{array}\right.
$$

We get:

$(3 ;-7 ; 1)$

## Problem Kuznetsov Analytic Geometry 14-28