This solution will guide you through finding the equation of a plane passing through three given points and then using the distance formula to find the unknown parameters.

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1. Key Concepts and Formulas

   • Equation of a Plane through Three Non-Collinear Points: The equation of a plane passing through three points $P_1(x_1, y_1, z_1)$, $P_2(x_2, y_2, z_2)$, and $P_3(x_3, y_3, z_3)$ can be found by first forming two vectors lying in the plane (e.g., $\overrightarrow{P_1P_2}$ and $\overrightarrow{P_1P_3}$). Their cross product, $\vec{n} = \overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3}$, gives the normal vector to the plane. If $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$, the plane's equation is $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$, which simplifies to $Ax+By+Cz=D$.
   • Distance of a Point from a Plane: The perpendicular distance $d$ of a point $(x_0, y_0, z_0)$ from a plane $Ax+By+Cz+D'=0$ is given by the formula: $d = \frac{|Ax_0+By_0+Cz_0+D'|}{\sqrt{A^2+B^2+C^2}}$ Note that if the plane equation is written as $Ax+By+Cz=D$, then $D'$ becomes $-D$, so the formula is $d = \frac{|Ax_0+By_0+Cz_0-D|}{\sqrt{A^2+B^2+C^2}}$. Consistency in the form of the plane equation is crucial.

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1. Step-by-Step Solution

   Step 1: Find the Equation of the Plane P

   We are given three points: $P_1(5,3,0)$, $P_2(13,3,-2)$, and $P_3(1,6,2)$. Our goal is to find the equation of the plane $P$ in the form $Ax+By+Cz+D'=0$.

   • Step 1.1: Form two vectors lying in the plane. We'll use $\overrightarrow{P_1P_2}$ and $\overrightarrow{P_1P_3}$ as they both originate from $P_1$ and lie within the plane. $\overrightarrow{P_1P_2} = (13-5)\hat{i} + (3-3)\hat{j} + (-2-0)\hat{k} = 8\hat{i} + 0\hat{j} - 2\hat{k}$ $\overrightarrow{P_1P_3} = (1-5)\hat{i} + (6-3)\hat{j} + (2-0)\hat{k} = -4\hat{i} + 3\hat{j} + 2\hat{k}$

   • Step 1.2: Calculate the normal vector to the plane. The normal vector $\vec{n}$ is perpendicular to all vectors in the plane, so we find it using the cross product of $\overrightarrow{P_1P_2}$ and $\overrightarrow{P_1P_3}$. $\vec{n} = \overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 8 & 0 & -2 \\ -4 & 3 & 2 \end{array}\right|$ Expanding the determinant: $\vec{n} = \hat{i}((0)(2) - (-2)(3)) - \hat{j}((8)(2) - (-2)(-4)) + \hat{k}((8)(3) - (0)(-4))$ $\vec{n} = \hat{i}(0+6) - \hat{j}(16-8) + \hat{k}(24-0)$ $\vec{n} = 6\hat{i} - 8\hat{j} + 24\hat{k}$ The direction ratios of the normal are $(6, -8, 24)$. We can simplify these by dividing by their common factor 2, giving $(3, -4, 12)$. This means $A=3, B=-4, C=12$.

   • Step 1.3: Determine the constant term of the plane equation. The preliminary equation of the plane is $3x - 4y + 12z = D$. To find $D$, we substitute the coordinates of any of the given points into this equation. Let's use $P_1(5,3,0)$: $3(5) - 4(3) + 12(0) = D$ $15 - 12 + 0 = D$ $D = 3$ So, the equation of the plane P is $3x - 4y + 12z = 3$. This can be written as $3x - 4y + 12z - 3 = 0$. Here, $A=3, B=-4, C=12, D'=-3$. The magnitude of the normal vector, $\sqrt{A^2+B^2+C^2}$, which forms the denominator in the distance formula, is: $\sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9+16+144} = \sqrt{169} = 13$

   Step 2: Calculate the Value of $\alpha$ for Point A

   Point A is $(3,4,\alpha)$, and its distance from plane P is given as 2. Using the distance formula $d = \frac{|Ax_0+By_0+Cz_0+D'|}{\sqrt{A^2+B^2+C^2}}$ with $A=3, B=-4, C=12, D'=-3$ and $(x_0,y_0,z_0) = (3,4,\alpha)$: $2 = \frac{|3(3) - 4(4) + 12(\alpha) - 3|}{13}$ $2 \times 13 = |9 - 16 + 12\alpha - 3|$ $26 = |12\alpha - 10|$ Solving the absolute value equation:

   • Case 1: $12\alpha - 10 = 26$ $12\alpha = 36$ $\alpha = 3$
   • Case 2: $12\alpha - 10 = -26$ $12\alpha = -16$ $\alpha = -\frac{16}{12} = -\frac{4}{3}$

   The problem states that $\alpha \in \mathbb{N}$ (natural numbers), so we choose the positive integer value: $\alpha = 3$

   Step 3: Calculate the Positive Value of $a$ for Point B

   Point B is $(2, \alpha, a)$, and its distance from plane P is given as 3. Substitute the value $\alpha=3$ into the coordinates of point B, making it $(2,3,a)$. Using the distance formula with $A=3, B=-4, C=12, D'=-3$ and $(x_0,y_0,z_0) = (2,3,a)$: $3 = \frac{|3(2) - 4(3) + 12(a) - 3|}{13}$ $3 \times 13 = |6 - 12 + 12a - 3|$ $39 = |12a - 9|$ Solving the absolute value equation:

   • Case 1: $12a - 9 = 39$ $12a = 48$ $a = \frac{48}{12} = 4$
   • Case 2: $12a - 9 = -39$ $12a = -30$ $a = -\frac{30}{12} = -\frac{5}{2}$

   The question asks for the positive value of $a$. From our two solutions, the positive value is: $a = 4$

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1. Common Mistakes & Tips

   • Cross Product Arithmetic: Be meticulous with signs and calculations when computing the cross product. A small error here propagates through the entire problem.
   • Absolute Value Interpretation: Always remember that $|X|=k$ implies two possibilities: $X=k$ and $X=-k$. Failing to consider both can lead to missing valid solutions.
   • Domain Restrictions: Pay close attention to conditions like $\alpha \in \mathbb{N}$ (natural numbers) or "positive value of $a$". These constraints help in selecting the correct answer from multiple mathematical solutions.

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1. Summary

   We began by determining the equation of plane P. We formed two vectors from the given points, calculated their cross product to find the normal vector, and then used one of the points to find the constant term. The simplified plane equation was $3x - 4y + 12z - 3 = 0$. Next, we used the distance formula for point A $(3,4,\alpha)$, given its distance from plane P is 2. This yielded $\alpha=3$ (as $\alpha$ must be a natural number). Finally, we applied the distance formula for point B $(2,3,a)$, given its distance from plane P is 3, which led to the positive value $a=4$.

The final answer is $\boxed{4}$, which corresponds to option (B).