Answer
Vector form:$\begin{bmatrix}{x \\y \\z}\end{bmatrix} = s\begin{bmatrix}{2 \\1\\2}\end{bmatrix}+t\begin{bmatrix}{-3 \\2\\1}\end{bmatrix}$
Parametric form: $x=2s-3t, y=s+2t,z=2s+t$
Work Step by Step
The vector form of a line is: $x=p+su+tv$
This implies, $\begin{bmatrix}{x \\y \\z}\end{bmatrix} = \begin{bmatrix}{0 \\0 \\0}\end{bmatrix}+s\begin{bmatrix}{2 \\1\\2}\end{bmatrix}+t\begin{bmatrix}{-3 \\2\\1}\end{bmatrix}$
or, $\begin{bmatrix}{x \\y \\z}\end{bmatrix} = s\begin{bmatrix}{2 \\1\\2}\end{bmatrix}+t\begin{bmatrix}{-3 \\2\\1}\end{bmatrix}$
Parametric equations of a line are defined as such equations which correspond to the components of the vector.
Thus, the parametric form of the equation of a line is:
$x=2s-3t, y=s+2t,z=2s+t$
Hence, the vector form is:$\begin{bmatrix}{x \\y \\z}\end{bmatrix} = s\begin{bmatrix}{2 \\1\\2}\end{bmatrix}+t\begin{bmatrix}{-3 \\2\\1}\end{bmatrix}$
The parametric form is: $x=2s-3t, y=s+2t,z=2s+t$
