The plane has intercepts
y = z = 0 ⇒ 3x = 6 ⇒ x = 2
x = z = 0 ⇒ 2y = 6 ⇒ y = 3
x = y = 0 ⇒ z = 6
and thus passes through the points (2, 0, 0), (0, 3, 0), and (0, 0, 6).
Parameterize the portion of the plane (call it P) by the vector function,
[tex]\vec r(s,t) = \left\langle 2(1-s)(1-t), 3s(1-t), 6t \right\rangle[/tex]
where 0 ≤ s ≤ 1 and 0 ≤ t ≤ 1.
Compute the normal vector to P :
[tex]\vec n = \dfrac{\partial \vec r}{\partial t} \times \dfrac{\partial \vec r}{\partial s} = -\left\langle 18-18t, 12-12t, 6-6t \right\rangle[/tex]
Then the flux of F through P is given by the surface integral,
[tex]\displaystyle \iint_P \vec F \cdot d\vec S = \int_0^1 \int_0^1 \left\langle 6s(1-s)(1-t)^2, 90st(1-t), 72t(1-s)(1-t) \right\rangle \cdot \vec n \, ds \, dt \\\\ = - \int_0^1 \int_0^1 \left(108(t-1)^3 s^2 + 108(t-1)^2(5t+1) s + 432t(t-1)^2\right) \, ds \, dt \\\\ = - \int_0^1 18(t-1)^2(41t+1) \, dt \\\\ = \boxed{-\frac{135}2}[/tex]
