## Calculus (3rd Edition)

$$x+2y-z=1.$$
We have $$\overrightarrow{PQ}=Q-P=(0,1,1)-(1,0,0)=\langle -1,1,1\rangle,$$ $$\overrightarrow{PR}=R-P=(2,0,1)-(1,0,0)=\langle 1,0,1\rangle .$$ Now, the normal vector $n$ is given by $$\overrightarrow{PQ} \times \overrightarrow{PR}= \left|\begin{array}{rrr}{i} & {j} & {k} \\ {-1} & {1} & {1} \\ {1} & {0} & {1}\end{array}\right|=i\left|\begin{array}{rrr} {1} & {1} \\ {0} & {1}\end{array}\right|-j\left|\begin{array}{rrr} {-1} & {1} \\ {1} & {1}\end{array}\right|+k\left|\begin{array}{rrr} {-1} & {1} \\ {1} & {0} \end{array}\right|\\ =i(1-0)-j(-1-1)+k(0-1)\\ =i+2j-k.$$ Now, using any of the given points, the equation of the plane is given by $$(x-1)+2(y-0)-(z-0)=0\Longrightarrow x+2y-z-1=0$$ Hence the equation is given by $$x+2y-z=1.$$