The plane has intercepts at (1, 0, 0), (0, 2, 0), and (0, 0, 6), so we can parameterize the surface [tex]\mathcal S[/tex] by
[tex]\mathbf r(u,v)=((1,0,0)(1-u)+(0,2,0)u)(1-v)+(0,0,6)v[/tex]
[tex]\mathbf r(u,v)=((1-u)(1-v),2u(1-v),6v)[/tex]
where [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Now
[tex]\|\mathbf r_u\times\mathbf r_v\|=2\sqrt{46}(1-v)[/tex]
so the surface integral reduces to
[tex]\displaystyle\iint_{\mathcal S}x\,\mathrm dS=2\sqrt{46}\int_{v=0}^{v=1}\int_{u=0}^{u=1}(1-u)(1-v)^2\,\mathrm du\,\mathrm dv=\frac{\sqrt{46}}3[/tex]
