Find the plane that goes through the points $\left(3, 0, -4\right)$, $\left(-2, -2, 5\right)$, and $\left(-1, 8, -3\right)$.

\begin{align*}

P &= \left(3, 0, -4\right) && \text{Define points}\\

Q &= \left(-2, -2, 5\right)\\

R &= \left(-1, 8, -3\right)\\\\

\overrightarrow{PQ} &= \left\langle -5, -2, 9 \right\rangle && \text{Vector through points}\\

\overrightarrow{PR} &= \left\langle -4, 8, 1 \right\rangle\\\\

\overrightarrow{PQ} \times \overrightarrow{PR} &= \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{-5}&{-2}&{9}\\{-4}&{8}&{1}\end{array}} \right| && \text{Take the cross product}\\

\overrightarrow{PQ} \times \overrightarrow{PR} &= \left| {\begin{array}{*{20}{c}}{-2}&{9}\\{8}&{1}\end{array}} \right|\vec i - \left| {\begin{array}{*{20}{c}}{-5}&{9}\\{-4}&{1}\end{array}} \right|\vec j + \left| {\begin{array}{*{20}{c}}{-5}&{-2}\\{-4}&{8}\end{array}} \right|\vec k\\\\

\overrightarrow{PQ} \times \overrightarrow{PR} &= \langle -74, -31, -48 \rangle && \text{Normal vector}\\

P &= \left(3, 0, -4\right) && \text{Point on plane}\\\\

0 &= \vec{n}\cdot (r - r_0) && \text{Equation of a plane}\\

&&&\text{$\vec n$: the normal vector}\\

&&&\text{$r_0$: the point on the plane} \\\\

0 &= \langle -74, -31, -48 \rangle \cdot \left(\langle x, y, z\rangle - \langle 3,0,5 \right\rangle) && \text{Equation of the plane}\\

0 &= -74(x-3) - 31y - 48(z-4) && \text{Simplified to scalar notation}

\end{align*}