Answer
$$6\sqrt 6-2 \sqrt 2$$
Work Step by Step
Apply cylindrical coordinates. $ x=r \cos \theta ;\\ y= r \sin \theta ;\\ z \gt 0$
We know that $ r(r, \theta)=xi+yj+zk $ or, $ r^2=x^2+y^2+z^2$
$ x^2-2y-2z=0$
$$ Surface \space Area =\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} \space dA\\=\int_{0}^{2} \int_{0}^{3x}\sqrt {x^2+2} \space dx \space dy\\ =\int_{0}^{2} 3x \sqrt {x^2+2} \space dx \\=6\sqrt 6-2 \sqrt 2$$
