To solve this problem step by step, here’s the breakdown:

Step 1: Define the function $f(x, y) = d^2$

The distance from a point on the plane $x - 2y + 3z = 12$ to the point $(0, 2, 2)$ is $d = \sqrt{(x - 0)^2 + (y - 2)^2 + (z - 2)^2}$. Squaring the distance gives: $f(x, y) = d^2 = x^2 + (y - 2)^2 + (z - 2)^2$ Substituting $z$ from the plane equation $z = \frac{12 - x + 2y}{3}$: $f(x, y) = x^2 + (y - 2)^2 + \left(\frac{12 - x + 2y}{3} - 2\right)^2$

Step 2: Simplify the function $f(x, y)$

Simplify the third term: $z - 2 = \frac{12 - x + 2y}{3} - 2 = \frac{12 - x + 2y - 6}{3} = \frac{6 - x + 2y}{3}$ Squaring this term: $\left(\frac{6 - x + 2y}{3}\right)^2 = \frac{(6 - x + 2y)^2}{9}$ Thus, $f(x, y)$ becomes: $f(x, y) = x^2 + (y - 2)^2 + \frac{(6 - x + 2y)^2}{9}$

Step 3: Find $f_x(x, y)$ and $f_y(x, y)$

To find critical points, compute partial derivatives.

Partial derivative with respect to $x$:

$f_x(x, y) = 2x + \frac{2}{9}(6 - x + 2y)(-1)$ Simplify: $f_x(x, y) = 2x - \frac{2}{9}(6 - x + 2y) = 2x - \frac{12 - 2x + 4y}{9}$ Combine terms: $f_x(x, y) = 2x - \frac{12}{9} + \frac{2x}{9} - \frac{4y}{9} = \frac{18x + 2x}{9} - \frac{12}{9} - \frac{4y}{9}$ $f_x(x, y) = \frac{20x - 4y - 12}{9}$

Partial derivative with respect to $y$:

$f_y(x, y) = 2(y - 2) + \frac{2}{9}(6 - x + 2y)(2)$ Simplify: $f_y(x, y) = 2y - 4 + \frac{4(6 - x + 2y)}{9} = 2y - 4 + \frac{24 - 4x + 8y}{9}$ Combine terms: $f_y(x, y) = \frac{18y + 8y}{9} - 4 + \frac{24 - 4x}{9} = \frac{26y - 4x + 24}{9} - 4$ Simplify further: $f_y(x, y) = \frac{26y - 4x + 24}{9} - \frac{36}{9} = \frac{26y - 4x - 12}{9}$

Step 4: Solve for the critical point

Set $f_x(x, y) = 0$ and $f_y(x, y) = 0$: $\frac{20x - 4y - 12}{9} = 0 \quad \text{and} \quad \frac{26y - 4x - 12}{9} = 0$ Simplify the first equation: $20x - 4y - 12 = 0 \quad \Rightarrow \quad 20x = 4y + 12 \quad \Rightarrow \quad 5x = y + 3 \quad \Rightarrow \quad y = 5x - 3$ Substitute $y = 5x - 3$ into the second equation: $26(5x - 3) - 4x - 12 = 0$ Simplify: $130x - 78 - 4x - 12 = 0 \quad \Rightarrow \quad 126x - 90 = 0 \quad \Rightarrow \quad x = \frac{90}{126} = \frac{5}{7}$ Solve for $y$: $y = 5\left(\frac{5}{7}\right) - 3 = \frac{25}{7} - 3 = \frac{25}{7} - \frac{21}{7} = \frac{4}{7}$ Thus, the critical point is: $(x, y) = \left(\frac{5}{7}, \frac{4}{7}\right)$

Step 5: Find the closest point on the plane

Substitute $x = \frac{5}{7}$ and $y = \frac{4}{7}$ into the plane equation $x - 2y + 3z = 12$: $\frac{5}{7} - 2\left(\frac{4}{7}\right) + 3z = 12 \quad \Rightarrow \quad \frac{5}{7} - \frac{8}{7} + 3z = 12$ $-\frac{3}{7} + 3z = 12 \quad \Rightarrow \quad 3z = 12 + \frac{3}{7} = \frac{84}{7} + \frac{3}{7} = \frac{87}{7}$ $z = \frac{87}{21} = \frac{29}{7}$ The closest point is: $\left(x, y, z\right) = \left(\frac{5}{7}, \frac{4}{7}, \frac{29}{7}\right)$

Let me know if you would like further clarifications!

Expanding Questions:

1. How do partial derivatives help find critical points in optimization problems?
2. Can this approach be generalized to higher dimensions?
3. What role does squaring the distance play in simplifying calculations?
4. How does substituting variables from constraints aid in solving optimization problems?
5. What other techniques exist for minimizing distances in constrained systems?

Tip: Always check whether the critical points lie within the domain of the problem!