This problem involves finding the distance from a point to a line in 3D space. The line's definition is indirect, requiring us to first determine its direction vector from the intersection of two planes.

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1. Key Concepts and Formulas

• Direction Vector of a Line of Intersection of Two Planes: If a line is formed by the intersection of two planes, $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$, its direction vector $\vec{d}$ is perpendicular to the normal vectors of both planes. Thus, $\vec{d}$ can be found by taking the cross product of their normal vectors: $\vec{d} = \vec{n_1} \times \vec{n_2}$, where $\vec{n_1} = (A_1, B_1, C_1)$ and $\vec{n_2} = (A_2, B_2, C_2)$.
• Equation of a Line in 3D (Symmetric Form): A line passing through a point $P_0(x_0, y_0, z_0)$ and having a direction vector $\vec{d} = (a,b,c)$ can be represented as: $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = \lambda$ where $\lambda$ is a scalar parameter. Any point on the line can be expressed parametrically as $(x_0+a\lambda, y_0+b\lambda, z_0+c\lambda)$.
• Distance of a Point from a Line (Vector Method): The shortest (perpendicular) distance from a point $Q$ to a line $L$ passing through a point $P$ with direction vector $\vec{d}$ is given by the formula: $\alpha = \frac{|\vec{PQ} \times \vec{d}|}{|\vec{d}|}$ Alternatively, one can find the foot of the perpendicular $R$ from $Q$ to $L$. If $R$ is a general point on $L$, then $\vec{QR}$ must be perpendicular to $\vec{d}$, meaning $\vec{QR} \cdot \vec{d} = 0$. Once $R$ is found, the distance is $|\vec{QR}|$.

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2. Step-by-Step Solution

Step 1: Determine the Direction Vector of Line $L$

• What we are doing: We need to find the direction vector of line $L$. The problem states that line $L$ is parallel to the line of intersection of two planes.
• Why this step is taken: Parallel lines share the same direction vector. Therefore, finding the direction vector of the line of intersection will give us the direction vector for line $L$. The direction vector of the line of intersection of two planes is found by taking the cross product of their normal vectors.

The given planes are:

1. $x+3y-2z-2=0$
2. $x-y+2z=0$

The normal vector to the first plane is $\vec{n_1} = (1, 3, -2)$. The normal vector to the second plane is $\vec{n_2} = (1, -1, 2)$.

The direction vector $\vec{d}$ of the line of intersection (and thus of line $L$) is found by taking the cross product of $\vec{n_1}$ and $\vec{n_2}$: $\vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ 1 & -1 & 2 \end{vmatrix}$ Expanding the determinant: $\vec{d} = \hat{i}((3)(2) - (-2)(-1)) - \hat{j}((1)(2) - (-2)(1)) + \hat{k}((1)(-1) - (3)(1))$ $\vec{d} = \hat{i}(6 - 2) - \hat{j}(2 + 2) + \hat{k}(-1 - 3)$ $\vec{d} = 4\hat{i} - 4\hat{j} - 4\hat{k}$ We can simplify this direction vector by dividing by the common factor of 4. This gives us a simpler, but equivalent, direction vector: $\vec{d} = (1, -1, -1)$

Step 2: Write the Equation of Line $L$

• What we are doing: We are writing the equation of line $L$.
• Why this step is taken: The equation of the line allows us to represent any point on the line using a single parameter, which is crucial for finding the foot of the perpendicular or applying the distance formula.

Line $L$ passes through the point $P(2,3,1)$ and has the direction vector $\vec{d} = (1, -1, -1)$. Using the symmetric form of the line equation: $L: \frac{x-2}{1} = \frac{y-3}{-1} = \frac{z-1}{-1} = \lambda$ From this, we can express the coordinates of any general point $R$ on line $L$ in terms of $\lambda$: $R(\lambda+2, -\lambda+3, -\lambda+1)$

Step 3: Calculate the Distance $\alpha$ using the Foot of the Perpendicular Method

• What we are doing: We are finding the shortest distance from the given point $Q(5,3,8)$ to line $L$.
• Why this step is taken: The shortest distance from a point to a line is always along the perpendicular from the point to the line. By finding the foot of the perpendicular, we can directly calculate this shortest distance.

Let the given point be $Q(5,3,8)$. Let $R(\lambda+2, -\lambda+3, -\lambda+1)$ be the foot of the perpendicular from $Q$ to $L$. First, form the vector $\vec{QR}$: $\vec{QR} = R - Q = ((\lambda+2)-5, (-\lambda+3)-3, (-\lambda+1)-8)$ $\vec{QR} = (\lambda-3, -\lambda, -\lambda-7)$ Since $\vec{QR}$ must be perpendicular to the direction vector $\vec{d} = (1, -1, -1)$, their dot product must be zero: $\vec{QR} \cdot \vec{d} = 0$ $(\lambda-3)(1) + (-\lambda)(-1) + (-\lambda-7)(-1) = 0$ $\lambda - 3 + \lambda + \lambda + 7 = 0$ $3\lambda + 4 = 0$ $\lambda = -\frac{4}{3}$ This value of $\lambda$ corresponds to the specific point $R$ on line $L$ that is the foot of the perpendicular from point $Q$.

Now, substitute the value of $\lambda = -\frac{4}{3}$ back into the expression for $\vec{QR}$: $\vec{QR} = \left(-\frac{4}{3}-3, -\left(-\frac{4}{3}\right), -\left(-\frac{4}{3}\right)-7\right)$ $\vec{QR} = \left(-\frac{4}{3}-\frac{9}{3}, \frac{4}{3}, \frac{4}{3}-\frac{21}{3}\right)$ $\vec{QR} = \left(-\frac{13}{3}, \frac{4}{3}, -\frac{17}{3}\right)$ The distance $\alpha$ is the magnitude of this vector $\vec{QR}$: $\alpha = |\vec{QR}| = \sqrt{\left(-\frac{13}{3}\right)^2 + \left(\frac{4}{3}\right)^2 + \left(-\frac{17}{3}\right)^2}$ $\alpha = \sqrt{\frac{169}{9} + \frac{16}{9} + \frac{289}{9}}$ $\alpha = \sqrt{\frac{169 + 16 + 289}{9}}$ $\alpha = \sqrt{\frac{474}{9}}$ $\alpha = \frac{\sqrt{474}}{3}$

Step 4: Alternative Method for Distance $\alpha$ (Vector Cross Product Formula)

• What we are doing: We are calculating the distance using the formula $\alpha = \frac{|\vec{PQ} \times \vec{d}|}{|\vec{d}|}$ to cross-check our result.
• Why this step is taken: Using an alternative method helps confirm the accuracy of our previous calculations.

Let $P(2,3,1)$ be a point on line $L$, and $Q(5,3,8)$ be the external point. The direction vector of line $L$ is $\vec{d} = (1, -1, -1)$. First, find the vector $\vec{PQ}$: $\vec{PQ} = Q - P = (5-2, 3-3, 8-1) = (3,0,7)$ Next, calculate the cross product $\vec{PQ} \times \vec{d}$: $\vec{PQ} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & 7 \\ 1 & -1 & -1 \end{vmatrix}$ $= \hat{i}(0 \cdot (-1) - 7 \cdot (-1)) - \hat{j}(3 \cdot (-1) - 7 \cdot 1) + \hat{k}(3 \cdot (-1) - 0 \cdot 1)$ $= \hat{i}(0 + 7) - \hat{j}(-3 - 7) + \hat{k}(-3 - 0)$ $= 7\hat{i} + 10\hat{j} - 3\hat{k} = (7, 10, -3)$ Now, find the magnitude of $\vec{PQ} \times \vec{d}$: $|\vec{PQ} \times \vec{d}| = \sqrt{7^2 + 10^2 + (-3)^2} = \sqrt{49 + 100 + 9} = \sqrt{158}$ Find the magnitude of the direction vector $\vec{d}$: $|\vec{d}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{1+1+1} = \sqrt{3}$ Finally, calculate the distance $\alpha$: $\alpha = \frac{|\vec{PQ} \times \vec{d}|}{|\vec{d}|} = \frac{\sqrt{158}}{\sqrt{3}}$ To compare with the previous method, we can write $\alpha = \frac{\sqrt{158}}{\sqrt{3}} = \frac{\sqrt{158} \times \sqrt{3}}{3} = \frac{\sqrt{474}}{3}$. Both methods yield the same distance.

Step 5: Final Calculation of $3\alpha^2$

• What we are doing: We are calculating the value $3\alpha^2$ as required by the problem.
• Why this step is taken: This is the final step to answer the question.

We have $\alpha = \frac{\sqrt{474}}{3}$. $3\alpha^2 = 3 \times \left(\frac{\sqrt{474}}{3}\right)^2$ $3\alpha^2 = 3 \times \frac{474}{9}$ $3\alpha^2 = \frac{474}{3}$ $3\alpha^2 = 158$

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3. Common Mistakes & Tips

• Cross Product Errors: Be careful with signs and order of operations when calculating the cross product of vectors. A common mistake is forgetting the negative sign for the $\hat{j}$ component.
• Simplifying Direction Vectors: Always simplify direction vectors by dividing by common factors. This makes subsequent calculations (like dot products and magnitudes) much easier and less prone to errors.
• Arithmetic Errors: Double-check all arithmetic, especially when dealing with fractions and square roots.
• Choosing the Right Method: Both the foot of the perpendicular method and the vector cross product formula are valid. Choose the one you are most comfortable with, but it's good practice to be familiar with both for verification.

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4. Summary

This problem required a multi-step approach in 3D geometry. We first determined the direction vector of the line $L$ by finding the cross product of the normal vectors of the two planes whose intersection formed a line parallel to $L$. Then, using the point $P(2,3,1)$ and the direction vector, we established the equation of line $L$. Finally, we calculated the perpendicular distance $\alpha$ from the point $(5,3,8)$ to line $L$ using two methods: by finding the foot of the perpendicular and by using the vector cross product formula. Both methods consistently yielded $\alpha = \frac{\sqrt{474}}{3}$. The final value of $3\alpha^2$ was then calculated.

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5. Final Answer

The final answer is $\boxed{158}$.