## Problem Statement

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

$$
\begin{aligned}
& \frac{x-1}{2}=\frac{y-2}{0}=\frac{z-4}{1} \\
& x-2 y+4 z-19=0
\end{aligned}
$$
## Solution

Let's write the parametric equations of the line.

$$
\frac{x-1}{2}=\frac{y-2}{0}=\frac{z-4}{1}=t \Rightarrow
$$

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

Substitute into the equation of the plane:

$(1+2 t)-2 \cdot 2+4(4+t)-19=0$

$1+2 t-4+16+4 t-19=0$

$6 t-6=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=1+2 \cdot 1=3 \\
y=2 \\
z=4+1=5
\end{array}\right.
$$

We get:

$(3 ; 2 ; 5)$

## Problem Kuznetsov Analytical Geometry 14-8