I assume you mean the plane [tex]x+2y+3z=1[/tex]. Its area over the region
[tex]R = \left\{(x,y) ~:~ x^2 + y^2 \le 6\right\}[/tex]
is given by the integral
[tex]\displaystyle \iint_R dA = \iint_R \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx \, dy[/tex]
where [tex]z=f(x,y) = \frac{1 - x - 2y}3[/tex].
We have
[tex]\dfrac{\partial f}{\partial x} = -\dfrac13[/tex]
[tex]\dfrac{\partial f}{\partial y} = -\dfrac23[/tex]
so that the area element is
[tex]dA = \sqrt{1 + \left(-\dfrac13\right)^2 + \left(-\dfrac23\right)^2} \,dx\,dy = \dfrac{\sqrt{14}}3\,dx\,dy[/tex]
Then we have
[tex]\displaystyle \iint_R dA = \frac{\sqrt{14}}3 \iint_R dx \, dy[/tex]
and the remaining integral is exactly the area of the disk [tex]x^2+y^2\le6[/tex]. Its radius is √6, so its area is π (√6)² = 6π. So the area of the surface is
[tex]\displaystyle \iint_R dA = \frac{\sqrt{14}}3 \cdot 6\pi = \boxed{2\sqrt{14}\pi}[/tex]
