## Problem Statement

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

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

$3 x+4 y+7 z-16=0$
## Solution

Let's write the parametric equations of the line.

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

Substitute into the equation of the plane:

$3(3+2 t)+4(-1+3 t)+7(-3+2 t)-16=0$

$9+6 t-4+12 t-21+14 t-16=0$

$32 t-32=0$

$t=1$

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

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

We get:

$(5 ; 2 ;-1)$

## Problem Kuznetsov Analytic Geometry 14-18