Find scalar equations of the form $\displaystyle ax + by + cz = 0$ for the following plane in $\displaystyle R^3$:

The plane through the points:

- $\displaystyle P(0;1;1)$
- $\displaystyle Q(1;0;1)$
- $\displaystyle R(1;1;0)$

$\displaystyle {PQ}^{\rightarrow} = (1;-1;0)$

$\displaystyle {QR}^{\rightarrow} = (0;1;-1)$

The perpendicular vector $\displaystyle = {PQ}^{\rightarrow} \times {QR}^{\rightarrow}$

Then I just picked point $\displaystyle Q$ to substitute into the equation:

$\displaystyle 1(x -1) + 1(y - 0) + 1(z - 1) = 0$

Multiplying it out:

$\displaystyle x + y + z = 2$

And that should be the equation of the plane?