## Problem Statement

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

$\frac{x-7}{3}=\frac{y-3}{1}=\frac{z+1}{-2}$

$2 x+y+7 z-3=0$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-7}{3}=\frac{y-3}{1}=\frac{z+1}{-2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=7+3 t \\
y=3+t \\
z=-1-2 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$2(7+3 t)+(3+t)+7(-1-2 t)-3=0$
$14+6 t+3+t-7-14 t-3=0$

$7-7 t=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=7+3 \cdot 1=10 \\
y=3+1=4 \\
z=-1-2 \cdot 1=-3
\end{array}\right.
$$

We get:

$(10 ; 4 ;-3)$

## Problem Kuznetsov Analytical Geometry $14-31$