you will receive a question-answer pair and be asked to rewrite just  the answer and explains your reasoning step by step, incorporating dynamic Chain of Thought (CoT), reflection, and verbal reinforcement learning. Follow these instructions:

1. Enclose all thoughts within <thinking> tags, exploring multiple angles and approaches.
2. Break down the solution into clear steps, providing a title and content for each step.
3. After each step, decide if you need another step or if you're ready to give the final answer.
4. Continuously adjust your reasoning based on intermediate results and reflections, adapting your strategy as you progress.
5. Regularly evaluate your progress, being critical and honest about your reasoning process.
6. Assign a confidence score between 0.0 and 1.0 to guide your approach:
   - 0.8+: Continue current approach
   - 0.5-0.7: Consider minor adjustments
   - Below 0.5: Seriously consider backtracking and trying a different approach
7. If unsure or if your score is low, backtrack and try a different approach, explaining your decision.
8. For mathematical problems, show all work explicitly using LaTeX for formal notation and provide detailed proofs.
9. Explore multiple solutions individually if possible, comparing approaches in your reflections.
10. Use your thoughts as a scratchpad, writing out all calculations and reasoning explicitly.
11. Use at least 5 methods to derive the answer and consider alternative viewpoints.
12. Be aware of your limitations as an AI and what you can and cannot do.

### Example Problem:
*Problem*: Solve for \( x \) in the equation \( 2x + 3 = 11 \).

### Expected Format:
*Step 1*: Subtract 3 from both sides of the equation.  
<thinking>Subtracting 3 from both sides should isolate the term with \( x \). Am I confident this is the correct first step? Yes, because it simplifies the equation to \( 2x = 8 \).</thinking>  
*Step 1 Confidence Score*: 0.85

*Step 2*: Divide both sides by 2.  
<thinking>Dividing by 2 will solve for \( x \). This is a straightforward division. Is there any reason to doubt this step? No, it should give \( x = 4 \).</thinking>  
*Step 2 Confidence Score*: 0.90

*Step 3*: Verify the solution by substituting \( x = 4 \) back into the original equation.  
<thinking>Substituting \( x = 4 \) back into the equation should confirm the solution. Does \( 2(4) + 3 = 11 \)? Yes, it does, so the solution is verified.</thinking>  
*Step 3 Confidence Score*: 1.00

Question: Find the rectangular coordinates of the point on the surface of the unit sphere that is farthest from the point $(2,3,5)$ in rectangular coordinates.

Let's break down the problem:

* The original problem involves converting a point from spherical coordinates to rectangular coordinates.
* In this new problem, we're not given the spherical coordinates of the point. Instead, we're asked to find the rectangular coordinates of a point on the surface of the unit sphere.
* The point on the unit sphere that is farthest from a given point $(2,3,5)$ can be found by maximizing the distance function.
* To solve this problem, we need to find the point on the unit sphere that maximizes the distance from $(2,3,5)$.
* We can use optimization techniques to find this point.

Let's assume the point on the unit sphere is $(x,y,z)$. We need to maximize the distance from $(2,3,5)$, which is given by the function $d(x,y,z) = \sqrt{(x-2)^2 + (y-3)^2 + (z-5)^2}$, subject to the constraint that the point is on the unit sphere, i.e., $x^2 + y^2 + z^2 = 1$.

Using the method of Lagrange multipliers, we can form the Lagrangian function

$$L(x,y,z,\lambda) = d(x,y,z) - \lambda (x^2 + y^2 + z^2 - 1)$$

Taking partial derivatives of $L$ with respect to $x$, $y$, $z$, and $\lambda$, we get:

$$\frac{\partial L}{\partial x} = \frac{x-2}{d} - 2\lambda x = 0$$
$$\frac{\partial L}{\partial y} = \frac{y-3}{d} - 2\lambda y = 0$$
$$\frac{\partial L}{\partial z} = \frac{z-5}{d} - 2\lambda z = 0$$
$$\frac{\partial L}{\partial \lambda} = x^2 + y^2 + z^2 - 1 = 0$$

Solving this system of equations will give us the values of $x$, $y$, and $z$ that maximize the distance from $(2,3,5)$.
Answer: The Lagrange multiplier method involves setting the gradient of the function equal to the gradient of the constraint, scaled by the Lagrange multiplier $\lambda$. Since the constraint is $x^2 + y^2 + z^2 = 1$, we have $\nabla (x^2 + y^2 + z^2) = (2x, 2y, 2z)$. The gradient of the distance function $d(x,y,z) = \sqrt{(x-2)^2 + (y-3)^2 + (z-5)^2}$ is $\nabla d = \left(\frac{x-2}{d}, \frac{y-3}{d}, \frac{z-5}{d}\right)$.

Setting these gradients equal, we get:
\[ (2x, 2y, 2z) = \lambda \left(\frac{x-2}{d}, \frac{y-3}{d}, \frac{z-5}{d}\right) \]

From the equality of the vectors, we can equate the corresponding components to get the system of equations:
\[ 2x = \lambda \frac{x-2}{d} \Rightarrow \lambda = \frac{2xd}{x-2} \]
\[ 2y = \lambda \frac{y-3}{d} \Rightarrow \lambda = \frac{2yd}{y-3} \]
\[ 2z = \lambda \frac{z-5}{d} \Rightarrow \lambda = \frac{2zd}{z-5} \]

Since $\lambda$ is the same in all equations, we can set them equal to each other:
\[ \frac{2xd}{x-2} = \frac{2yd}{y-3} = \frac{2zd}{z-5} \]

Since $d$ is the same in all equations, we can cancel it out:
\[ \frac{x}{x-2} = \frac{y}{y-3} = \frac{z}{z-5} \]

Now, we can solve this system of equations to find the point $(x,y,z)$ on the unit sphere.

We can rewrite the equations as:
\[ x(y-3) = y(x-2) \]
\[ x(z-5) = z(x-2) \]

Simplifying these equations, we get:
\[ xy - 3x = xy - 2y \Rightarrow 2y = 3x \]
\[ xz - 5x = xz - 2z \Rightarrow 2z = 5x \]

Now, we can use the constraint $x^2 + y^2 + z^2 = 1$ to find the values of $x$, $y$, and $z$:
\[ x^2 + \left(\frac{3x}{2}\right)^2 + \left(\frac{5x}{2}\right)^2 = 1 \]
\[ x^2 + \frac{9x^2}{4} + \frac{25x^2}{4} = 1 \]
\[ 4x^2 + 9x^2 + 25x^2 = 4 \]
\[ 38x^2 = 4 \]
\[ x^2 = \frac{4}{38} = \frac{2}{19} \]
\[ x = \pm \sqrt{\frac{2}{19}} \]

Since $y = \frac{3x}{2}$ and $z = \frac{5x}{2}$, we have:
\[ y = \pm \frac{3}{2} \sqrt{\frac{2}{19}} = \pm \frac{3}{\sqrt{38}} \]
\[ z = \pm \frac{5}{2} \sqrt{\frac{2}{19}} = \pm \frac{5}{\sqrt{38}} \]

The point $(x,y,z)$ on the unit sphere that is farthest from $(2,3,5)$ is:
\[ \boxed{\left(-\frac{2}{\sqrt{38}}, -\frac{3}{\sqrt{38}}, -\frac{5}{\sqrt{38}}\right)} \]