\left.\begin{array}{l}{[\text { Method of coordinates in space }} \\ \text { [ Parametric equations of a line }]\end{array}\right]

Given points $A(-3 ; 0 ; 1)$ and $D(1 ; 3 ; 2)$. Write the parametric equations of the line $A D$.
Point $M(x ; y ; z)$ lies on the line $A D$ if and only if the vector $\overrightarrow{A M}=(x+3 ; y ; z-1)$ is collinear with the vector

$$
\overrightarrow{A D}=(1-(-3) ; 3-0 ; 2-1)=(4 ; 3 ; 1)
$$

The vector $\overrightarrow{A M}$ is collinear with the vector $\overrightarrow{A D}$ if and only if $\overrightarrow{A M}=t \cdot \overrightarrow{A D}$. The latter equality is equivalent to

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

## Answer

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