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Step 1: Express Each Line in Vector Form
The given lines in symmetric form are:
Line 1:
$ \frac{x - \lambda}{0} = \frac{y - 3}{4} = \frac{z + 6}{1}, $
Line 2:
$ \frac{x + \lambda}{3} = \frac{y}{-4} = \frac{z - 6}{0}. $
From these forms, identify:
• A point on Line 1 as
$ P_{1}(\lambda,\;3,\;-6), $
• A direction vector for Line 1 as
$ \vec{d}_{1} = \langle 0,\;4,\;1\rangle. $
• A point on Line 2 as
$ P_{2}(-\lambda,\;0,\;6), $
• A direction vector for Line 2 as
$ \vec{d}_{2} = \langle 3,\;-4,\;0\rangle. $
Step 2: Form the Vector Connecting a Point on Line 1 to a Point on Line 2
Consider the vector from $P_{1}$ to $P_{2}$:
$
\vec{c} = P_{2} - P_{1}
= \langle -\lambda - \lambda,\;0 - 3,\;6 - (-6)\rangle
= \langle -2\lambda,\;-3,\;12\rangle.
$
Step 3: Find the Cross Product of the Direction Vectors
A vector perpendicular to both $ \vec{d}_{1} $ and $ \vec{d}_{2} $ is given by
$
\vec{n} = \vec{d}_{1} \times \vec{d}_{2}.
$
Compute the determinant:
$
\vec{n}
=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 4 & 1 \\
3 & -4 & 0
\end{vmatrix}.
$
Expand:
$
\vec{n}
= \mathbf{i}\,(4 \cdot 0 - 1 \cdot (-4))
\;-\; \mathbf{j}\,(0 \cdot 0 - 1 \cdot 3)
\;+\; \mathbf{k}\,(0 \cdot (-4) - 4 \cdot 3).
$
Simplify each component:
$
\vec{n} = \langle 4,\;3,\;-12\rangle.
$
Step 4: Use the Formula for the Shortest Distance Between Two Skew Lines
The shortest distance $ d $ between the two skew lines is given by the absolute value of the scalar projection of $ \vec{c} $ onto the direction of $ \vec{n} $:
$
d = \left|\frac{\vec{c} \cdot \vec{n}}{\|\vec{n}\|}\right|.
$
First compute
$
\vec{c} \cdot \vec{n}.
$
$
\vec{c} \cdot \vec{n}
= \langle -2\lambda,\;-3,\;12\rangle \cdot \langle 4,\;3,\;-12\rangle
= (-2\lambda)\times 4 + (-3)\times 3 + (12)\times (-12).
$
$
\vec{c} \cdot \vec{n} = -8\lambda - 9 - 144
= -8\lambda - 153.
$
Next, find the magnitude of $ \vec{n} $:
$
\|\vec{n}\| = \sqrt{4^2 + 3^2 + (-12)^2}
= \sqrt{16 + 9 + 144}
= \sqrt{169}
= 13.
$
Hence,
$
d = \left|\frac{-8\lambda - 153}{13}\right|.
$
Step 5: Apply the Given Distance Condition
It is given that the shortest distance $ d = 13. $
So,
$
13 = \left|\frac{-8\lambda - 153}{13}\right|.
$
Multiply both sides by 13:
$
169 = \bigl|-8\lambda - 153\bigr|.
$
This splits into two cases:
1) $ -8\lambda - 153 = 169, $
2) $ -8\lambda - 153 = -169. $
Step 6: Solve for $ \lambda $ in Each Case
Case 1:
$
-8\lambda - 153 = 169
\;\;\Rightarrow\;\; -8\lambda = 169 + 153 = 322
\;\;\Rightarrow\;\; \lambda = -\frac{322}{8} = -\frac{161}{4}.
$
Case 2:
$
-8\lambda - 153 = -169
\;\;\Rightarrow\;\; -8\lambda = -169 + 153 = -16
\;\;\Rightarrow\;\; \lambda = 2.
$
Step 7: Calculate the Required Expression
We have two valid values of $ \lambda $:
$
S = \left\{-\frac{161}{4},\,2\right\}.
$
We want
$
8 \left|\sum_{\lambda \in S} \lambda\right|.
$
First sum the solutions:
$
\lambda_{1} + \lambda_{2}
= \left(-\frac{161}{4}\right) + 2
= 2 - \frac{161}{4}
= \frac{8}{4} - \frac{161}{4}
= -\frac{153}{4}.
$
Its absolute value is
$
\left|-\frac{153}{4}\right|
= \frac{153}{4}.
$
Now multiply by 8:
$
8 \times \frac{153}{4}
= 2 \times 153
= 306.
$
Step 8: Final Answer
Therefore, $ 8 \left|\sum_{\lambda \in S} \lambda\right| = 306. $
