Let's solve the problem step by step to find the point \(P(x_0, y_0, z_0)\) on the surface \(z = 3y^2 + 2x^2\) that is closest to the plane \(z = y + x - 4\).

### Step 1: Define the distance between a point and the plane

The distance between a point \( (x_0, y_0, z_0) \) and the plane \( z = y + x - 4 \) is given by:

\[
\text{Distance} = \frac{|z_0 - y_0 - x_0 + 4|}{\sqrt{1^2 + 1^2 + (-1)^2}} = \frac{|z_0 - y_0 - x_0 + 4|}{\sqrt{3}}
\]

### Step 2: Use Lagrange multipliers to minimize the distance

We want to minimize the distance function \( D^2(x, y, z) = (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 \), but since \( z_0 \) lies on the surface \( z = 3y^2 + 2x^2 \), we minimize:

\[
f(x, y, z) = z - 3y^2 - 2x^2 = 0
\]
under the constraint:
\[
g(x, y, z) = z - y - x + 4 = 0
\]

### Step 3: Gradient calculations

The gradients are:
\[
\nabla f = (-4x, -6y, 1)
\]
\[
\nabla g = (-1, -1, 1)
\]

Setting \(\nabla f = \lambda \nabla g\):
\[
(-4x, -6y, 1) = \lambda (-1, -1, 1)
\]

This gives us the system:
1. \( -4x = -\lambda \)
2. \( -6y = -\lambda \)
3. \( 1 = \lambda \)

From \( 1 = \lambda \), we have \( \lambda = 1 \). So:
\[
x = \frac{1}{4}, \quad y = \frac{1}{6}
\]

### Step 4: Substitute into the surface equation

Substitute \( x = \frac{1}{4} \) and \( y = \frac{1}{6} \) into the surface equation to find \( z_0 \):

\[
z_0 = 3 \left(\frac{1}{6}\right)^2 + 2 \left(\frac{1}{4}\right)^2 = 3 \times \frac{1}{36} + 2 \times \frac{1}{16}
\]
\[
z_0 = \frac{1}{12} + \frac{1}{8} = \frac{2}{24} + \frac{3}{24} = \frac{5}{24}
\]

Thus, the point \( P(x_0, y_0, z_0) = \left(\frac{1}{4}, \frac{1}{6}, \frac{5}{24}\right) \).

### Step 5: Calculate the distance to the plane

Finally, the distance from this point to the plane \( z = y + x - 4 \) is:

\[
\text{Distance} = \frac{\left|\frac{5}{24} - \frac{1}{6} - \frac{1}{4} + 4\right|}{\sqrt{3}}
\]

Simplifying:
\[
\frac{5}{24} - \frac{1}{6} - \frac{1}{4} + 4 = \frac{5}{24} - \frac{4}{24} - \frac{6}{24} + \frac{96}{24} = \frac{91}{24}
\]

So the distance is:
\[
\frac{\frac{91}{24}}{\sqrt{3}} = \frac{91}{24\sqrt{3}} \approx 1.10
\]

---

### Final Answer:
The point closest to the plane is \( P\left(\frac{1}{4}, \frac{1}{6}, \frac{5}{24}\right) \), and the distance between this point and the plane is approximately 1.10 units.

---

Would you like more details, or have any questions? 

Here are some related questions you might find interesting:

1. How to find the closest point using optimization techniques?
2. What is the significance of using the Lagrange multiplier method?
3. How does the distance formula for a point to a plane work?
4. What are the alternative methods to solve such a problem?
5. How does the constraint affect the solution in optimization?
6. How can we interpret the gradients in this context?
7. Can we use any software tools to solve such problems more easily?
8. How would the solution change if the surface or the plane was different?

**Tip:** Lagrange multipliers are a powerful tool in optimization, especially when dealing with constraints. Understanding how to apply them in different contexts can greatly expand your problem-solving abilities in mathematics.

Learn how to find the closest point on a given surface to a plane using advanced optimization techniques, specifically Lagrange multipliers. Detailed step-by-step solution provided with calculations and explanations.

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