The area is 97.211 sq units.
This part of the plane is a triangle. We can Call it δ. We can find the intercepts by setting two variables to 0 simultaneously; we'd find, for instance, that y=z=0 means 3x=15 i.e., x=3 , so that (3, 0, 0) is one vertex of the triangle. Similarly, we'd find that (0, 5, 0) and (0, 0, 15) are the other two vertices.
Next, we can parameterize the surface by
s(u,v)=[tex]\int\limits^a_b {3(1-u)(1-v),5u(1-v),15v} \, dx[/tex]
so surface element is
dS= IIsu*svII =15[tex]\sqrt{42}[/tex](1-v)dudv
Then the area of is given by the surface integral
[tex]\int\limits^a_b {} \, \int\limits^a_b {dS} \, dx = 15\sqrt[n]{42} \int\limits^a_b {x} \, dx \int\limits^a_b {1-v} \, dudv[/tex]=97.211
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