Answer
$$2x + y -2z =-2$$
Work Step by Step
Write the general equation of a plane in $R^3$ passing through the point $P=(x_0,y_0, z_0)$ along with the normal vector $n=\lt p,q,r\gt$ such that:
$$p(x - x_0) + q(y - y_0) + r(z - z_0) = 0~~~~(1)$$
Here, we have $n=\lt 1,0, 1\gt \times \lt 0, 2,1\gt=-2i-j+2k$ or, $n=\lt -2, -1, 2 \gt$
Plug the values into the equation (1) to obtain:
$$ ( -2 )(x - 1) -1(y -2) + 2 (z-3) = 0 $$
After simplifying, we get:
$$-2x-y+2z-2 =0\\ 2x + y -2z =-2$$
