Answer
Vector form:$\begin{bmatrix}{x \\y \\z}\end{bmatrix} = \begin{bmatrix}{6 \\-4 \\-3}\end{bmatrix}+s\begin{bmatrix}{0 \\1\\1}\end{bmatrix}+t\begin{bmatrix}{-1 \\1\\1}\end{bmatrix}$
Parametric form: $x=6-t, y=-4+s+t,z=-3+s+t$
Work Step by Step
The vector form of a plane is: $x=p+su+tv$
This implies, $\begin{bmatrix}{x \\y \\z}\end{bmatrix} = \begin{bmatrix}{6 \\-4 \\-3}\end{bmatrix}+s\begin{bmatrix}{0 \\1\\1}\end{bmatrix}+t\begin{bmatrix}{-1 \\1\\1}\end{bmatrix}$
Parametric equations of a plane are defined as such equations which correspond to the components of the vector.
Thus, the parametric form of the equation of a plane is:
$x=6-t, y=-4+s+t,z=-3+s+t$
Hence, the vector form is:$\begin{bmatrix}{x \\y \\z}\end{bmatrix} = \begin{bmatrix}{6 \\-4 \\-3}\end{bmatrix}+s\begin{bmatrix}{0 \\1\\1}\end{bmatrix}+t\begin{bmatrix}{-1 \\1\\1}\end{bmatrix}$
The parametric form is: $x=6-t, y=-4+s+t,z=-3+s+t$
