Find the general equation of the plane containing the points $$P_{1}=(1,-1,2)\quad P_{2}=( 3,0,0)\quad P_{3}=(4,2,1)$$

Solution The vectors $\mathbf{v}=\overrightarrow{P_{1}~P_{2}}=2 \mathbf{i}+\mathbf{j}-2\mathbf{k}$ and $\mathbf{w}=\overrightarrow{P_{1}~P_{3} }=3\mathbf{i}+3\mathbf{j}-\mathbf{k}$ lie in the plane. The vector $$\begin{equation*} \mathbf{N}=\mathbf{v}\times \mathbf{w}=\left\vert \begin{array}{r@{\quad}r@{\quad}r} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] 2 & 1 & -2 \\[3pt] 3 & 3 & -1 \end{array} \right\vert =\left\vert \begin{array}{r@{\quad}r} 1 & -2 \\[3pt] 3 & -1 \end{array} \right\vert \mathbf{i}-\left\vert \begin{array}{r@{\quad}r} 2 & -2 \\[3pt] 3 & -1 \end{array} \right\vert \mathbf{j}+\left\vert \begin{array}{r@{\quad}r} 2 & 1 \\[3pt] 3 & 3 \end{array} \right\vert \mathbf{k}=5\mathbf{i}-4\mathbf{j}+3\mathbf{k} \end{equation*}$$

is orthogonal to both $\mathbf{v}$ and $\mathbf{w}$ and so is normal to the plane. Using the point $(1,-1,2)$ and the vector $\mathbf{N}$, the general equation of the plane is $$\begin{eqnarray*} 5(x-1)-4(y+1)+3(z-2)& =&0 \\[2pt] 5x-5-4y-4+3z-6& =&0 \\[2pt] 5x-4y+3z& =&15 \end{eqnarray*}$$