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

Given points are (2, 1, 0), (3, -2, -2), and (3, 1, 7).

We know, equation of a line passing through 3 non-collinear points (x1 ,   y1 , z1 ), (x2 , y2 , z2 ) and (x3 , y3 , z3 ) is given as:

$\begin{vmatrix} x-x_{1} &y-y_{1} &z-z_{1} \\ x_{2}-x_{1}&y_{2}-y_{1} &z_{2}-z_{1} \\ x_{3}-x_{1}&y_{3}-y_{1} &z_{3}-z_{1} \end{vmatrix}=0$

Where, (x1 ,   y1 , z1 ) = (2, 1, 0)

(x2 , y2 , z2 ) = (3, -2, -2)

(x3 , y3 , z3 ) = (3, 1, 7)

Therefore, x1 = 2, y1 = 1, z1 = 0; x2 = 3, y2 = -2, z2 = -2; x3 = 3, y3 = 1, z3 = 7

Substituting these values in the line equation,

$\begin{vmatrix} x-2 &y-1 &z-0 \\ 3-2&-2-1 &-2-0 \\ 3-2&1-1 &7-0 \end{vmatrix}=0\\ \\ \\ \begin{vmatrix} x-2 &y-1 &z-0 \\ 1&-3 &-2 \\ 1&0 &7 \end{vmatrix}=0$

$\\ \\ \begin{vmatrix} x-2 &y-1 &z-0 \\ 1&-3 &-2 \\ 1&0 &7 \end{vmatrix}=\left ( x-2 \right )\left ( \left ( -3 \times 7 \right )-\left ( -2 \times 0 \right ) \right )$

$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=\left ( x-2 \right )\left ( -21-0 \right )-\left ( y-1 \right )\left ( 7-(-2) \right )+z\left ( 0-(-3) \right )$

$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=\left ( x-2 \right )\left ( -21 \right )-\left ( y-1 \right )\left ( 7+2 \right )+z\left ( 0+3 \right )$

$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=-21\left ( x-2 \right )-9\left ( y-1 \right )+3z$

$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=-21x+42-9y+9+3z$

$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=-21x-9y+3z+42+9$

$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=-21x-9y+3z+51$

Now, since
$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=0$

$\Rightarrow -21x -9y + 3z + 51 = 0\\ \Rightarrow -21x - 9y + 3z = -51\\ \Rightarrow -3(7x + 3y - z) = -3 \times 17\\ \Rightarrow 7x + 3y - z = 17$

Hence, the required equation of the plane is 7x + 3y - z = 17.