Answer
$ 3 \sqrt {14}$
Work Step by Step
We have $A(S)= \iint_{D} \sqrt {1+(-3)^2+(-2)^2 } dA$
and $A(S)=\sqrt {14} \iint_{D} dA .....(a)$
and $\iint_{D} dA$ is the area of the region $D$
Now, the area of the triangle with vertices $A(2,0), B(0,3), C(0,0) $ can be found as: $\iint_{D} dA=\dfrac{1}{2} |AC \times BC|=\dfrac{1}{2} |\lt 2-0, 0-0 \gt \times \lt 0-0, 3-0 \gt |=\dfrac{1}{2} |2i \times 3 j|=3$
Thus, $\iint_{D} dA= 3 ....(b)$
From equations (a) and (b) we can conclude that the area of the plane in first quadrant becomes: $ 3 \sqrt {14}$
