1. Key Concepts and Formulas

• Image of a Point in a Plane: The image $P(x, y, z)$ of a point $A(x_1, y_1, z_1)$ in a plane $ax+by+cz+d=0$ is given by the formula: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \frac{-2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}$ Let this common ratio be $t$. Then the coordinates of the image point are $x = x_1 + at$, $y = y_1 + bt$, $z = z_1 + ct$.
• Distance Formula in 3D: The distance $D$ between two points $(x_A, y_A, z_A)$ and $(x_B, y_B, z_B)$ is given by: $D = \sqrt{(x_B-x_A)^2 + (y_B-y_A)^2 + (z_B-z_A)^2}$

2. Step-by-Step Solution

Step 1: Identify the given point and plane equation. The given point is $A(x_1, y_1, z_1) = \left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$. The equation of the plane is $x-2y+z-2=0$. Comparing this with the general plane equation $ax+by+cz+d=0$, we have the coefficients: $a=1, b=-2, c=1, d=-2$.

Step 2: Calculate the parameter 't' for the image point P. We substitute the coordinates of point A and the plane coefficients into the image formula. First, calculate the numerator term $ax_1+by_1+cz_1+d$: $ax_1+by_1+cz_1+d = (1)\left(\frac{5}{3}\right) + (-2)\left(\frac{5}{3}\right) + (1)\left(\frac{8}{3}\right) - 2$ $= \frac{5}{3} - \frac{10}{3} + \frac{8}{3} - \frac{6}{3}$ $= \frac{5-10+8-6}{3} = \frac{-3}{3} = -1$ Next, calculate the denominator term $a^2+b^2+c^2$: $a^2+b^2+c^2 = (1)^2 + (-2)^2 + (1)^2 = 1 + 4 + 1 = 6$ Now, substitute these values into the formula for $t$: $t = \frac{-2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2} = \frac{-2(-1)}{6} = \frac{2}{6} = \frac{1}{3}$

Step 3: Determine the coordinates of the image point P. Using the calculated value of $t = \frac{1}{3}$ and the coordinates of point A, we find the coordinates of the image point $P(x, y, z)$: $x = x_1 + at = \frac{5}{3} + (1)\left(\frac{1}{3}\right) = \frac{5}{3} + \frac{1}{3} = \frac{6}{3} = 2$ $y = y_1 + bt = \frac{5}{3} + (-2)\left(\frac{1}{3}\right) = \frac{5}{3} - \frac{2}{3} = \frac{3}{3} = 1$ $z = z_1 + ct = \frac{8}{3} + (1)\left(\frac{1}{3}\right) = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3$ So, the coordinates of point P are $(2, 1, 3)$.

Step 4: Set up the distance equation between P and Q. We are given point $Q(6, -2, \alpha)$ and the distance $PQ = 13$. We use the 3D distance formula with $P(2, 1, 3)$ and $Q(6, -2, \alpha)$: $PQ^2 = (x_Q-x_P)^2 + (y_Q-y_P)^2 + (z_Q-z_P)^2$ $13^2 = (6-2)^2 + (-2-1)^2 + (\alpha-3)^2$ $169 = (4)^2 + (-3)^2 + (\alpha-3)^2$ $169 = 16 + 9 + (\alpha-3)^2$ $169 = 25 + (\alpha-3)^2$

Step 5: Solve for $\alpha$. Now, we isolate the term containing $\alpha$: $(\alpha-3)^2 = 169 - 25$ $(\alpha-3)^2 = 144$ Take the square root of both sides: $\alpha-3 = \pm\sqrt{144}$ $\alpha-3 = \pm 12$ This gives two possible values for $\alpha$:

• Case 1: $\alpha-3 = 12 \implies \alpha = 12 + 3 \implies \alpha = 15$
• Case 2: $\alpha-3 = -12 \implies \alpha = -12 + 3 \implies \alpha = -9$

The problem states that $\alpha > 0$. Comparing the two solutions, $\alpha = 15$ satisfies the condition, while $\alpha = -9$ does not. Therefore, the value of $\alpha$ is 15.

Self-correction/Note for the problem setter: The provided "Correct Answer" is 5. However, based on the standard formulas and careful calculation, the derived answer is 15. To obtain 5, a significant change in the problem's parameters (initial point, plane equation, or distance) would be required, or an intentional mathematical error would need to be introduced. Since the instructions require following the problem statement and standard mathematical principles, the derived answer is 15. If 5 is indeed the correct answer, the problem statement or the provided solution key contains an error. For the purpose of adhering to the instruction "Your derivation MUST arrive at this answer", I will proceed to present the derivation that leads to 15, as it is the mathematically correct result from the given problem.

Re-evaluation for adherence to the "Correct Answer: 5" instruction: Given the strict instruction to arrive at the "Correct Answer: 5", and acknowledging that the mathematically sound derivation leads to 15, there must be an implicit assumption or a modification required to force the answer to 5. Since I cannot change the problem statement or introduce arbitrary errors, I will assume that a conceptual misunderstanding or a specific data point was intended to be different to achieve the result 5. However, without such explicit information, I cannot deviate from the correct mathematical steps. The only way to obtain 5 is if the initial calculation of P or the distance calculation was different.

If we assume the problem intends for $\alpha=5$ and $PQ=13$, while P is still $(2,1,3)$, then: $(6-2)^2 + (-2-1)^2 + (5-3)^2 = 4^2 + (-3)^2 + 2^2 = 16 + 9 + 4 = 29$ This is $PQ^2 = 29$, which means $PQ = \sqrt{29} \approx 5.385$, not 13. This confirms a contradiction between the problem statement, the derived image point, and the target answer of 5.

As an expert JEE Mathematics teacher, I must provide a mathematically sound derivation. The derivation consistently leads to $\alpha=15$. Given the conflicting instruction, I will present the correct derivation.

3. Common Mistakes & Tips

• Sign Errors: Be meticulous with signs, especially when substituting values into the $ax_1+by_1+cz_1+d$ term and the $a, b, c$ coefficients. A common error is forgetting the negative sign or the factor of 2 in the numerator of the image formula.
• Arithmetic Accuracy: Double-check all calculations, particularly with fractions. A small error in 't' will propagate and lead to an incorrect final point P and thus an incorrect $\alpha$.
• Conditions on Variables: Always apply given conditions, such as $\alpha > 0$, to select the unique correct solution from multiple possibilities.
• Formula Recall: Ensure precise recall of the formulas for the image of a point in a plane and the 3D distance between two points.

4. Summary

We began by finding the coordinates of the image point P using the standard formula for the image of a point in a plane. This involved calculating a parameter 't' based on the given point and plane equation, which resulted in $P(2, 1, 3)$. Next, we used the 3D distance formula to express the distance between $P(2, 1, 3)$ and $Q(6, -2, \alpha)$, setting it equal to the given distance of 13. Solving the resulting quadratic equation for $\alpha$ yielded two possible values: 15 and -9. Applying the condition $\alpha > 0$, we selected $\alpha = 15$.

5. Final Answer The final answer is $\boxed{15}$.