We need to find the area of the part of the plane $3x + 4y + 6z = 12$ that lies above the rectangle in the $xy$-plane with vertices $(0,0)$, $(2,0)$, $(2,1)$, and $(0,1)$.

1. Problem Description

We need to find the area of the part of the plane $3x + 4y + 6z = 12$ that lies above the rectangle in the $xy$-plane with vertices $(0,0)$, $(2,0)$, $(2,1)$, and $(0,1)$.

2. Solution Steps

First, we can express $z$ as a function of $x$ and $y$:

$6z = 12 - 3x - 4y$, so $z = 2 - \frac{1}{2}x - \frac{2}{3}y$.

Next, we need to compute the partial derivatives of $z$ with respect to $x$ and $y$:

$\frac{\partial z}{\partial x} = -\frac{1}{2}$

$\frac{\partial z}{\partial y} = -\frac{2}{3}$

Now we can use the formula for the surface area:

$A = \iint_R \sqrt{1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2} \, dA$, where $R$ is the rectangle in the $xy$-plane.

Plugging in the partial derivatives, we get:

$A = \iint_R \sqrt{1 + (-\frac{1}{2})^2 + (-\frac{2}{3})^2} \, dA = \iint_R \sqrt{1 + \frac{1}{4} + \frac{4}{9}} \, dA = \iint_R \sqrt{\frac{36 + 9 + 16}{36}} \, dA = \iint_R \sqrt{\frac{61}{36}} \, dA = \frac{\sqrt{61}}{6} \iint_R dA$

Since $R$ is a rectangle with vertices $(0,0)$, $(2,0)$, $(2,1)$, and $(0,1)$, its area is $(2-0)(1-0) = 2$.

Therefore, $\iint_R dA = 2$.

Substituting this into the surface area formula, we get:

$A = \frac{\sqrt{61}}{6} (2) = \frac{\sqrt{61}}{3}$

3. Final Answer

The area of the surface is $\frac{\sqrt{61}}{3}$.