## Problem Statement

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

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

$4 x+2 y-z-11=0$
## Solution

Let's write the parametric equations of the line.

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

Substitute into the equation of the plane:

$4(1+2 t)+2(1-t)-(-2+3 t)-11=0$

$4+8 t+2-2 t+2-3 t-11=0$

$3 t-3=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=1+2 \cdot 1=3 \\
y=1-1=0 \\
z=-2+3 \cdot 1=1
\end{array}\right.
$$

We get:

$(3 ; 0 ; 1)$

## Problem Kuznetsov Analytic Geometry 14-11