Answer
$$ 3 \sqrt {14}$$
Work Step by Step
We know that $A(S)= \iint_{D} \sqrt {1+(-3)^2+(-2)^2 } dA$
or, $A(S)=\sqrt {14} \iint_{D} dA$
where $\iint_{D} dA$ defines the area of the region $D$ .
The area of the triangle with vertices $A(2,0), B(0,3), C(0,0) $ is:
$$\iint_{D} dA=\dfrac{1}{2} |AC \times BC|=\dfrac{1}{2} |\lt 2-0, 0-0 \gt \times \lt 0-0, 3-0 \gt | \\=\dfrac{1}{2} |2i \times 3 j| \\=3$$
So, the area of the plane in the first quadrant is:
$$A(S)=\sqrt {14} \iint_{D} dA= 3 \sqrt {14}$$
