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Step 1: Identify the Direction Vector of the Given Line
We are given that line L passes through the point P(2, 3, 1) and is parallel to the line whose equations can be written as
x + 3y - 2z - 2 = 0 and x - y + 2z = 0.
To find a direction vector for L, we solve these two equations simultaneously in a parametric sense to identify the direction ratios (1, -1, -1). Thus, the direction vector of the line L can be taken as
\langle 1, -1, -1 \rangle.
Step 2: Write the Parametric Form of Line L
Since L passes through P(2, 3, 1) and has direction vector \langle 1, -1, -1 \rangle , its parametric equations can be written as:
\displaystyle \frac{x - 2}{1} \;=\; \frac{y - 3}{-1} \;=\; \frac{z - 1}{-1} \;=\; \lambda.
Hence, any point on L can be written as:
B(\lambda + 2,\; -\lambda + 3,\; -\lambda + 1).
Step 3: Find the Foot of the Perpendicular from the Point (5, 3, 8) to Line L
Let A(5, 3, 8) be the external point. We want the perpendicular from A onto the line L. If B is the foot of the perpendicular on L, then the vector
\overrightarrow{AB} must be orthogonal (dot product zero) to the direction vector of L, \langle 1, -1, -1 \rangle.
First, write
\overrightarrow{AB} \;=\;
\bigl((\lambda + 2) - 5\bigr)\hat{i} \;+\;
\bigl((- \lambda + 3) - 3\bigr)\hat{j} \;+\;
\bigl((- \lambda + 1) - 8\bigr)\hat{k}.
This simplifies to:
\overrightarrow{AB} \;=\; (\lambda - 3)\,\hat{i} \;+\; (-\lambda)\,\hat{j} \;+\; (-\lambda - 7)\,\hat{k}.
Imposing the perpendicularity condition
\overrightarrow{AB}\cdot \langle 1, -1, -1 \rangle = 0
gives us:
(\lambda - 3)\cdot 1 \;+\; (-\lambda)\cdot (-1)\;+\; (-\lambda - 7)\cdot (-1) \;=\; 0.
So,
(\lambda - 3) + \lambda + (\lambda + 7) = 0 \quad\Longrightarrow\quad 3\lambda + 4 = 0 \quad\Longrightarrow\quad \lambda = -\frac{4}{3}.
Step 4: Compute the Coordinates of the Foot of Perpendicular
Plugging \lambda = -\frac{4}{3} into the parametric form of L:
B\Bigl(-\frac{4}{3} + 2,\; -(-\frac{4}{3}) + 3,\; -(-\frac{4}{3}) + 1\Bigr)
= \Bigl(\frac{6 - 4}{3}, \frac{9 + 4}{3}, \frac{3 + 4}{3}\Bigr)
= \Bigl(\frac{2}{3}, \frac{13}{3}, \frac{7}{3}\Bigr).
Step 5: Calculate the Distance \alpha
The distance \alpha from A to L is simply the length of the vector \overrightarrow{AB} . Now,
\overrightarrow{AB}
= \Bigl(\frac{2}{3} - 5,\; \frac{13}{3} - 3,\; \frac{7}{3} - 8\Bigr)
= \Bigl(-\frac{13}{3},\; \frac{4}{3},\; -\frac{17}{3}\Bigr).
Hence,
\displaystyle
|\overrightarrow{AB}|
= \sqrt{\Bigl(-\frac{13}{3}\Bigr)^2 + \Bigl(\frac{4}{3}\Bigr)^2 + \Bigl(-\frac{17}{3}\Bigr)^2}
= \sqrt{\frac{169}{9} + \frac{16}{9} + \frac{289}{9}}
= \sqrt{\frac{474}{9}}
= \frac{\sqrt{474}}{3}
= \alpha.
Step 6: Evaluate 3\alpha^2
We have \alpha = \frac{\sqrt{474}}{3}. Therefore,
\alpha^2 = \left(\frac{\sqrt{474}}{3}\right)^2 = \frac{474}{9}.
Thus,
3\,\alpha^2 = 3 \times \frac{474}{9} = \frac{3 \times 474}{9} = \frac{1422}{9} = 158.
So the value of 3\alpha^2 is 158.
