To determine the minimum distance from the origin $(0, 0, 0)$ to the plane $3x + 2y + z = 12$, we can use the method of Lagrange multipliers. This method involves minimizing the Euclidean distance function subject to a constraint.

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Step 1: Define the objective and constraint functions.

• Objective function: $D(x, y, z) = \sqrt{x^2 + y^2 + z^2}$, which is the Euclidean distance from the origin to a point $(x, y, z)$.
• Constraint: $g(x, y, z) = 3x + 2y + z - 12 = 0$.

To simplify calculations, we can minimize the square of the distance function: $f(x, y, z) = x^2 + y^2 + z^2.$

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Step 2: Apply the method of Lagrange multipliers.

The method introduces a multiplier $\lambda$ and requires that: $\nabla f = \lambda \nabla g,$ where $\nabla f$ and $\nabla g$ are the gradients of the objective and constraint functions, respectively.

• $\nabla f = \left( 2x, 2y, 2z \right)$.
• $\nabla g = \left( 3, 2, 1 \right)$.

This gives the system of equations: $2x = \lambda \cdot 3, \quad 2y = \lambda \cdot 2, \quad 2z = \lambda \cdot 1.$

Thus:

1. $2x = 3\lambda$ $\quad\Rightarrow\quad x = \frac{3\lambda}{2}$.
2. $2y = 2\lambda$ $\quad\Rightarrow\quad y = \lambda$.
3. $2z = \lambda$ $\quad\Rightarrow\quad z = \frac{\lambda}{2}$.

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Step 3: Use the constraint to solve for $\lambda$.

From the constraint $3x + 2y + z = 12$, substitute the expressions for $x$, $y$, and $z$ in terms of $\lambda$: $3\left(\frac{3\lambda}{2}\right) + 2(\lambda) + \frac{\lambda}{2} = 12.$ Simplify: $\frac{9\lambda}{2} + 2\lambda + \frac{\lambda}{2} = 12.$ $\frac{9\lambda + 4\lambda + \lambda}{2} = 12 \quad\Rightarrow\quad \frac{14\lambda}{2} = 12 \quad\Rightarrow\quad 7\lambda = 12 \quad\Rightarrow\quad \lambda = \frac{12}{7}.$

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Step 4: Find the coordinates $(x, y, z)$.

Now, substitute $\lambda = \frac{12}{7}$ into the expressions for $x$, $y$, and $z$: $x = \frac{3\lambda}{2} = \frac{3 \cdot \frac{12}{7}}{2} = \frac{36}{14} = \frac{18}{7},$ $y = \lambda = \frac{12}{7},$ $z = \frac{\lambda}{2} = \frac{\frac{12}{7}}{2} = \frac{6}{7}.$

Thus, the point on the plane closest to the origin is: $\left( \frac{18}{7}, \frac{12}{7}, \frac{6}{7} \right).$

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Step 5: Calculate the minimum distance.

The minimum distance is the Euclidean distance from the origin to the point $\left( \frac{18}{7}, \frac{12}{7}, \frac{6}{7} \right)$: $D = \sqrt{\left( \frac{18}{7} \right)^2 + \left( \frac{12}{7} \right)^2 + \left( \frac{6}{7} \right)^2}.$ $D = \sqrt{\frac{324}{49} + \frac{144}{49} + \frac{36}{49}} = \sqrt{\frac{504}{49}} = \sqrt{10.2857} \approx 3.2.$

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Final Answer:

The minimum distance from the origin to the plane $3x + 2y + z = 12$ is approximately: $D \approx 3.2.$

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Would you like me to elaborate on any step?

Here are five related questions you may find helpful:

1. What are other applications of Lagrange multipliers?
2. How would the solution change for a different plane equation?
3. Can we find the closest distance using projection techniques?
4. How do we apply Lagrange multipliers in higher dimensions?
5. What is the geometric interpretation of this minimum distance?

Tip: Squaring the distance function simplifies optimization problems because it removes the square root, making the equations easier to handle.