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We find the equation of a plane by taking the cross product of 2 vectors from the differences between 2 sets of points. So we start with our points. Let's make $A=(4,0,0)$, $B = (0,1,0)$, and $C=(0,0,2)$. If we take the differences between the points $A$ and $B$, and $B$ and $C$, we create vectors $\overrightarrow{AB} = \langle 4, -1, 0 \rangle$ and $\overrightarrow{BC} = \langle 0,1,-2\rangle$. We then take the cross product of these vectors $\overrightarrow{AB} \times \overrightarrow{BC}$ and get the matrix $$\begin{vmatrix} I & J & K\ 4 & -1 & 0\ 0 & 1& -2\ \end{vmatrix} $$ This matrix equals $2I -8J +5K$, and if we pick our favorite of the three original points, we can create an equation. I chose point $A$ and found the equation $$2(x -4) -8(y-0)+5(z-0) = 0$$ which, if we distribute, equals $$2x-8y+5z = 8$$ which we can easily use to find our integral. I hope this helps!

We find the equation of a plane by taking the cross product of 2 vectors from the differences between 2 sets of points. So we start with our points. Let's make $A=(4,0,0)$, $B = (0,1,0)$, and $C=(0,0,2)$. If we take the differences between the points $A$ and $B$, and $B$ and $C$, we create vectors $\overrightarrow{AB} = \langle 4, -1, 0 \rangle$ and $\overrightarrow{BC} = \langle 0,1,-2\rangle$. We then take the cross product of these vectors $\overrightarrow{AB} \times \overrightarrow{BC}$ and get the matrix $$\begin{vmatrix} I & J & K\ K\ 4 & -1 & 0\ 0 & 1& -2\ \end{vmatrix} $$ This matrix equals $2I -8J +5K$, and if we pick our favorite of the three original points, we can create an equation. I chose point $A$ and found the equation $$2(x -4) -8(y-0)+5(z-0) = 0$$ which, if we distribute, equals $$2x-8y+5z = 8$$ which we can easily use to find our integral. I hope this helps!