Parameterize [tex]T[/tex] by
[tex]\vec r(u,v)=(1-v)((1-u)(2\,\vec k)+u\,\vec\jmath)+v\,\vec\imath=v\,\vec\imath+u(1-v)\,\vec\jmath+2(1-u)(1-v)\,\vec k[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Take the normal vector to [tex]T[/tex] to be
[tex]\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=2(v-1)\,\vec\imath+2(v-1)\,\vec\jmath+(v-1)\,\vec k[/tex]
with magnitude
[tex]\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{(2(v-1))^2+(2(v-1))^2+(v-1)^2}=3(1-v)[/tex]
Then the integral is
[tex]\displaystyle\iint_Tyz\,\mathrm dS=\int_0^1\int_0^12u(1-u)(1-v)^2(3(1-v))\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle6\int_0^1\int_0^1u(1-u)(1-v)^3\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle6\left(\int_0^1u(1-u)\,\mathrm du\right)\left(\int_0^1(1-v)^3\,\mathrm dv\right)=\boxed{\frac14}[/tex]
