This part of the plane lies above a triangle with boundaries [tex]x=0[/tex] and [tex]y=0[/tex] along the coordinate axes, as well as the line
[tex]z=0 \implies 5x + 3y = 15 \implies y = \dfrac{15 - 5x}3[/tex]
When [tex]y=0[/tex], we have [tex]15-5x=0\implies x=3[/tex]. So this triangle is the set
[tex]T = \left\{(x,y)\in\Bbb R^2 ~:~ 0\le x\le3 \text{ and } 0\le y\le\dfrac{15-5x}3\right\}[/tex]
Also, when [tex]x=0[/tex], we have [tex]y=\frac{15}3=5[/tex]. So the triangle has length 3 and width 5, hence area 1/2•3•5 = 15/2.
Let [tex]z=f(x,y) = 15 - 5x - 3y[/tex]. Then the area of the plane over [tex]T[/tex] is
[tex]\displaystyle \iint_T dA = \iint_T \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx \, dy[/tex]
We have
[tex]\dfrac{\partial f}{\partial x} = -5 \implies \left(\dfrac{\partial f}{\partial x}\right)^2 = 25[/tex]
[tex]\dfrac{\partial f}{\partial y} = -3 \implies \left(\dfrac{\partial f}{\partial y}\right)^2 = 9[/tex]
[tex]\implies\displaystyle \iint_T dA = \sqrt{35} \iint_T dx\,dy = \boxed{\frac{15\sqrt{35}}2}[/tex]
since the integral
[tex]\displaystyle \iint_TdA[/tex]
is exactly the area of [tex]T[/tex], 15/2.
