Answer
$3 \sqrt {14}$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be computed as: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
$\iint_{D} dA$ is the projection of the surface on the $xy$-plane.
Area of the given surface can be computed as: $A(S)=\iint_{D} \sqrt {1+(-3)^2+(-2)^2} dA=\sqrt {14} \iint_{D} dA$
$\iint_{D} dA$ represents the area of the region inside $D$.
Now, the area of the triangle is : $=\dfrac{1}{2} \times (\ Base) \times (\ height) =\dfrac{1}{2} \times (2)(3)=3$
Then, we have: $A(S)=\sqrt {14} \iint_{D} dA=3 \sqrt {14}$
