Answer
$x+y+z=2$
Work Step by Step
The vector between $(0,1,1)$ and $(1,0,1)$ is $\lt1,-1,0\gt$.
The vector between $(1,1,0)$ and $(1,0,1)$ is $\lt0,-1,1\gt$.
After taking the cross product of the direction vectors, we get
$\lt1,1,1\gt $.
The general form of the equation of the plane is:
$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
or, $ax+by+cz=ax_0+by_0+cz_0$
Plugging in the given values, we get
$(x-0)+(y-1)+(z-1)=0$
After simplification, we get
$x+y+z=2$
