This part of the plane is a triangle. Call it [tex]\mathcal S[/tex]. We can find the intercepts by setting two variables to 0 simultaneously; we'd find, for instance, that [tex]y=z=0[/tex] means [tex]5x=20\implies x=4[/tex], so that (4, 0, 0) is one vertex of the triangle. Similarly, we'd find that (0, 5, 0) and (0, 0, 20) are the other two vertices.
Next, we can parameterize the surface by
[tex]\mathbf s(u,v)=\langle4(1-u)(1-v),5u(1-v),20v\rangle[/tex]
so that the surface element is
[tex]\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|=20\sqrt{42}(1-v)\,\mathrm du\,\mathrm dv[/tex]
Then the area of [tex]\mathcal S[/tex] is given by the surface integral
[tex]\displaystyle\iint_{\mathcal S}\mathrm dS=20\sqrt{42}\int_{u=0}^{u=1}\int_{v=0}^{v=1}(1-v)\,\mathrm dv\,\mathrm du[/tex]
[tex]\displaystyle=20\sqrt{42}\int_{v=0}^{v=1}(1-v)\,\mathrm dv=10\sqrt{42}\approx64.8074[/tex]
