Answer
\[3x-19y-2z=0\]
Work Step by Step
\[\begin{align}
& \text{Points }\underbrace{\left( 0,0,0 \right)}_{\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)},\left( 2,0,3 \right),\left( -3,-1,5 \right) \\
& \text{Let }\mathbf{u}\text{ be the vector from }\left( 0,0,0 \right)\text{ to }\left( 2,0,3 \right) \\
& \mathbf{u}=\left\langle 2-0,0-0,3-0 \right\rangle \\
& \mathbf{u}=\left\langle 2,0,3 \right\rangle \\
& \text{Let }\mathbf{v}\text{ be the vector from }\left( 0,0,0 \right)\text{ to }\left( -3,-1,5 \right) \\
& \mathbf{v}=\left\langle -3-0,-1-0,5-0 \right\rangle \\
& \mathbf{v}=\left\langle -3,-1,5 \right\rangle \\
& \mathbf{n}=\mathbf{u}\times \mathbf{v}=\left| \begin{matrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 0 & 3 \\
-3 & -1 & 5 \\
\end{matrix} \right| \\
& \mathbf{n}=\mathbf{u}\times \mathbf{v}=\left| \begin{matrix}
0 & 3 \\
-1 & 5 \\
\end{matrix} \right|\mathbf{i}-\left| \begin{matrix}
2 & 0 \\
-3 & -1 \\
\end{matrix} \right|\mathbf{j}+\left| \begin{matrix}
2 & 3 \\
-3 & 5 \\
\end{matrix} \right|\mathbf{k} \\
& \mathbf{n}=3\mathbf{i}-19\mathbf{j}-2\mathbf{k} \\
& \mathbf{n}=\left\langle a,b,c \right\rangle \\
& a=3,\text{ }b=-19,\text{ }c=-2 \\
& \text{The equation of the plane is given by} \\
& a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0 \\
& \text{Where }\underbrace{\left( 0,0,0 \right)}_{\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)},\text{ then} \\
& 3\left( x-0 \right)-19\left( y-0 \right)-2\left( z-0 \right)=0 \\
& 3x-19y-2z=0 \\
\end{align}\]
