Compute the flux of the vector field 
F = 

xy, 4yz, 3zx

 through the portion of the plane 
3x + 2y + z = 6
 in the first octant with the downward orientation.

To compute the flux of the vector field \( \mathbf{F} = (xy, 4yz, 3zx) \) through the portion of the plane \( 3x + 2y + z = 6 \) in the first octant, with a downward orientation, we'll follow these steps:

### Step 1: Parameterize the Surface
The equation of the plane is:
\[
3x + 2y + z = 6
\]
Solving for \( z \) in terms of \( x \) and \( y \):
\[
z = 6 - 3x - 2y
\]
This gives us a parameterization for \( z \) on the surface in terms of \( x \) and \( y \).

### Step 2: Find the Bounds in the First Octant
In the first octant, \( x, y, z \geq 0 \), so:
- When \( x = 0 \), \( z = 6 - 2y \), meaning \( y \leq 3 \).
- When \( y = 0 \), \( z = 6 - 3x \), meaning \( x \leq 2 \).

Thus, the region on the \( xy \)-plane is bounded by \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 3 - \frac{3}{2}x \).

### Step 3: Compute the Normal Vector
To find the flux, we need the normal vector. Since the surface is given by \( z = 6 - 3x - 2y \), let:
\[
\mathbf{r}(x, y) = (x, y, 6 - 3x - 2y)
\]
Then, the partial derivatives are:
\[
\frac{\partial \mathbf{r}}{\partial x} = (1, 0, -3) \quad \text{and} \quad \frac{\partial \mathbf{r}}{\partial y} = (0, 1, -2)
\]
The cross product \( \frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y} \) gives the normal vector:
\[
\mathbf{n} = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -3 \\ 0 & 1 & -2 \end{matrix} \right| = (3, 2, 1)
\]
Since the orientation is downward, we take \( \mathbf{n} = (-3, -2, -1) \).

### Step 4: Set Up the Flux Integral
The flux \( \iint_S \mathbf{F} \cdot d\mathbf{S} \) is given by:
\[
\iint_D \mathbf{F} \cdot \mathbf{n} \, dA
\]
where \( D \) is the projection of the surface onto the \( xy \)-plane, and \( d\mathbf{S} = \mathbf{n} \, dA \).

Substitute \( \mathbf{F} = (xy, 4yz, 3zx) \) and \( \mathbf{n} = (-3, -2, -1) \):
\[
\mathbf{F} \cdot \mathbf{n} = xy(-3) + 4y(6 - 3x - 2y)(-2) + 3x(6 - 3x - 2y)(-1)
\]

### Step 5: Simplify the Integral and Evaluate

Given the setup,

Compute the flux of the vector field F = (xy, 4yz, 3zx) through the portion of the plane 3x + 2y + z = 6 in the first octant with the downward orientation.

This solution shows how to compute the flux of the vector field F = (xy, 4yz, 3zx) through a plane in the first octant, using surface integrals and parameterization. The problem involves vector calculus techniques suitable for college-level calculus.

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