To find the surface area of the plane, we must find the partial derivative of [tex] f(x) = 9x + 2y + z -3 [/tex]. Then, we have
[tex] \frac {\partial f}{\partial x} = 9 [/tex]
[tex] \frac {\partial f}{\partial y} = 2 [/tex]
[tex] \frac {\partial f}{\partial z} = 1 [/tex]
We can compute for the surface area of the elliptical cylinder,
[tex] \frac{x^2}{81} + \frac{y^2}{4} = 1 [/tex] , with
[tex] Surface Area = \int \int\limits_C { \sqrt{ \frac{9^{2} + 2^{2} + 1^{2}}{1^{2}} } } \, dx \,dy [/tex]
knowing that we're working on the plane's surface from the cylinder,
[tex] Surface Area = \sqrt{86} \pi \int \int \,dx \,dy [/tex]
[tex] Surface Area = \sqrt{86} \pi (9-0)(2-0) [/tex]
[tex] Surface Area = 18 \sqrt{86} \pi [/tex]
Thus, the area around the plane of f(x) is [tex] 18 \sqrt{86} \pi [/tex].
Answer: [tex] 18 \sqrt{86} \pi [/tex] units²
