Find the equation of the plane passing through the points $(2, 1, 0)$,$(3, -2, -2)$,and $(3, 1, 7)$.

(N/A) The equation of a plane passing through three non-collinear points $(x_{1}, y_{1}, z_{1})$,$(x_{2}, y_{2}, z_{2})$,and $(x_{3}, y_{3}, z_{3})$ is given by the determinant equation:
$\begin{vmatrix} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix} = 0$
Substituting the given points $(2, 1, 0)$,$(3, -2, -2)$,and $(3, 1, 7)$:
$\begin{vmatrix} x-2 & y-1 & z-0 \\ 3-2 & -2-1 & -2-0 \\ 3-2 & 1-1 & 7-0 \end{vmatrix} = 0$
Simplifying the matrix:
$\begin{vmatrix} x-2 & y-1 & z \\ 1 & -3 & -2 \\ 1 & 0 & 7 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$(x-2)(-21 - 0) - (y-1)(7 - (-2)) + z(0 - (-3)) = 0$
$(x-2)(-21) - (y-1)(9) + z(3) = 0$
$-21x + 42 - 9y + 9 + 3z = 0$
$-21x - 9y + 3z + 51 = 0$
Dividing the entire equation by $-3$:
$7x + 3y - z = 17$
Thus,the required equation of the plane is $7x + 3y - z = 17$.