Let \(P=(2,0,0) Q=(0,4,0)\) and \(R=(0,0,2)\)

First, find a normal vector n

\(n=\vec{PQ}\cdot\vec{PR}\)

Therefore,

\(\vec{PQ}=(0,4,0)-(2,0,0)=(-2,4,0)\)

\(\vec{PR}=(0,0,2)-(2,0,0)=(-2,0,2)\)

\(n=\vec{PQ}\cdot\vec{PR}=\begin{bmatrix}i&j&k\\-2&4&0\\-2&0&2\end{bmatrix}\)

\(n=\begin{bmatrix}4&0\\0&2\end{bmatrix}i-\begin{bmatrix}-2&0\\-2&2\end{bmatrix}j+\begin{bmatrix}-2&4\\-2&0\end{bmatrix}k\)

\(n=(8-0)i-(-4+0)j+(0+8)k\)

\(n=8i+4j+8k\)

\(n=(8,4,8)\)

We have the equation \(8x+4y+8z=d\)

Now choose any one of the three points, using \(P=(2,0,0)\)

\(d=n\cdot\vec{OP}\)

\(d=(8,4,8)\cdot(2,0,0)\)

\(d=16\)

\(d=1\)

Therefore

\(8x+4y+8z=16\)

\(2x+y+2z=4\)

Results

\(2x+y+2z=4\)