## Calculus (3rd Edition)

$$2x+y+2z=4$$
We have $$\overrightarrow{PQ}=Q-P=(0,4,0)-(2,0,0)=\langle -2,4,0\rangle,$$ $$\overrightarrow{PR}=R-P=(0,0,2)-(2,0,0)=\langle -2,0,2\rangle .$$ Now, the normal vector $n$ is given by $$\overrightarrow{PQ} \times \overrightarrow{PR}= \left|\begin{array}{rrr}{i} & {j} & {k} \\ {-2} & {4} & {0} \\ {-2} & {0} & {2}\end{array}\right|=i\left|\begin{array}{rrr} {4} & {0} \\ {0} & {2}\end{array}\right|-j\left|\begin{array}{rrr} {-2} & {0} \\ {-2} & {2}\end{array}\right|+k\left|\begin{array}{rrr} {-2} & {4} \\ {-2} & {0} \end{array}\right|\\ =i(8-0)-j(-4-0)+k(0+8)\\ =8i-4j+8k.$$ Now, using any of the given points, the equation of the plane is given by $$8(x-2)+4(y-0)+8(z-0)=0\Longrightarrow 8x+4y+8z-16=0$$ Hence the equation is given by $$8x+4y+8z=16.$$ This can be simplified to (divide by $4$): $$2x+y+2z=4.$$