Find equation for a plane orthogonal to two other planes.

Question

How to find plane equation which goes through point

A(2,1,0)

and is orthogonal to the planes

x - y + z = 0 

and

2x - 3y + z - 1 = 0

Answer 1

Hint...the normal to the new plane is the cross product of the normals of the given planes, since it is orthogonal to these. Then use $$\underline{r}\cdot\underline{n}=\underline{a}\cdot\underline{n}$$

Answer 2

The equation of a plane, through point $(x_0,y_0,z_0)$ and normal vector $\mathbf{n}$ is $$ \mathbf{n} \cdot \left( {\mathbf{x} - \mathbf{x}_{\,0} } \right) = 0 $$ where clearly $\mathbf{x}=(x,y,z)$
So if you take $\mathbf{n} = \mathbf{u} \times \mathbf{v}$, i.e. $$ \mathbf{n} = \left( {1, - 1,1} \right) \times \left( {2, - 3,1} \right) = \left( {2,1, - 1} \right) $$ and plug it into the general equation above (with $\mathbf{x}_{\,0}$ having the coordinates of $A$).