Problem based on area projection in 3-D geometry Considering a Quadrilateral $ABCD$ where $A(0,0,0), B(2,0,2), C(2,2\sqrt 2,2), D(0,2\sqrt2,0)$. Basically I have to find the Area of projection of quadrilateral $ABCD$ on the plane $x+y-z=3$.
I have tried to first find the projection of the points $A,B,C,D$ individually on the plane and then using the projected points find the vectors $\vec{AB}$ and $\vec{BC}$ and then using $|\vec{AB}\cdot \vec{BC}|$ , but I was unable to find the projected points.
Is it the correct approach? If it is not I would highly appreciate a correct approach for the problem.

As indicated by Intelligenti pauca in the above comments, the best way to go is to find the area of the quadrilateral, then multiply the area found by the cosine of the angle between the two planes which is the same angle between the normals to the planes (or its supplement).
$\begin{equation} \begin{split}
\text{Area} &= \frac{1}{2} ( | AB \times AC | + | AC \times AD | ) \\
&= \frac{1}{2} ( | (2,0,2) \times (2, 2\sqrt{2}, 2) | + | (2, 2 \sqrt{2}, 2) \times (0, 2 \sqrt{2}, 0) | ) \\
&= 2 ( | (- \sqrt{2}, 0, \sqrt{2} ) | + | (-\sqrt{2},  0, \sqrt{2} ) | )\\
& = 2 ( 2 + 2 ) = 8 \end{split} \end{equation}$
The normal to the plane of the quadrilateral is along $(-1, 0, 1)$ and the normal to the projection plane is along $(1, 1, -1)$, therefore, if $\theta$ is the angle between these two normals, then,
$\cos \theta = \dfrac{ (-1, 0, 1) \cdot (1, 1, -1) }{\sqrt{2}\sqrt{3} } = -\sqrt{\dfrac{2}{3}} $
Hence, the projected area is equal to $8\sqrt{\dfrac{2}{3}}$
