## Problem Statement

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

$$
\begin{aligned}
& \frac{x-2}{-1}=\frac{y-3}{-1}=\frac{z+1}{4} \\
& x+2 y+3 z-14=0
\end{aligned}
$$
## Solution

Let's write the parametric equations of the line.

$$
\frac{x-2}{-1}=\frac{y-3}{-1}=\frac{z+1}{4}=t \Rightarrow
$$

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

Substitute into the equation of the plane:

$$
\begin{aligned}
& (2-t)+2(3-t)+3(-1+4 t)-14=0 \\
& 2-t+6-2 t-3+12 t-14=0 \\
& 9 t-9=0 \\
& t=1
\end{aligned}
$$

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

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

We get:

$(1 ; 2 ; 3)$

## Problem Kuznetsov Analytic Geometry 14-1