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Step-by-Step Solution
Step 1: Express each line in parametric form
Line 1 is given by:
\frac{x + \sqrt{6}}{2} \;=\; \frac{y - \sqrt{6}}{3} \;=\; \frac{z - \sqrt{6}}{4}.
Let this common ratio be t . Then the coordinates of any point on Line 1 can be written as:
x = 2t - \sqrt{6}, \quad y = 3t + \sqrt{6}, \quad z = 4t + \sqrt{6}.
Similarly, Line 2 is given by:
\frac{x - \lambda}{3} \;=\; \frac{y - 2\sqrt{6}}{4} \;=\; \frac{z + 2\sqrt{6}}{5}.
Let this common ratio be s . Then the coordinates of any point on Line 2 can be written as:
x = 3s + \lambda, \quad y = 4s + 2\sqrt{6}, \quad z = 5s - 2\sqrt{6}.
Step 2: Identify direction vectors and a point on each line
The direction vector of Line 1 is
\vec{d_1} = \langle 2,\, 3,\, 4 \rangle.
A convenient point on Line 1 (by setting t = 0 ) is
P_1 = (-\sqrt{6},\; \sqrt{6},\; \sqrt{6}).
The direction vector of Line 2 is
\vec{d_2} = \langle 3,\, 4,\, 5 \rangle.
A convenient point on Line 2 (by setting s = 0 ) is
P_2 = (\lambda,\; 2\sqrt{6},\; -2\sqrt{6}).
Step 3: Formula for the shortest distance between two skew lines
The shortest distance D between two skew lines with direction vectors \vec{d_1}, \vec{d_2} and points P_1, P_2 on them, respectively, is given by:
D \;=\; \frac{\bigl|(\overrightarrow{P_1P_2}) \cdot (\vec{d_1} \times \vec{d_2})\bigr|}{\bigl|\vec{d_1} \times \vec{d_2}\bigr|}.
Here,
\overrightarrow{P_1P_2} = P_2 - P_1 = \bigl(\lambda + \sqrt{6},\; 2\sqrt{6} - \sqrt{6},\; -2\sqrt{6} - \sqrt{6}\bigr) = \bigl(\lambda + \sqrt{6},\; \sqrt{6},\; -3\sqrt{6}\bigr).
Step 4: Compute the cross product of the direction vectors
We have
\vec{d_1} = \langle 2,\; 3,\; 4 \rangle, \quad \vec{d_2} = \langle 3,\; 4,\; 5 \rangle.
The cross product \vec{d_1} \times \vec{d_2} is given by the determinant:
\vec{d_1} \times \vec{d_2}
= \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 3 & 4 \\
3 & 4 & 5
\end{vmatrix}
= \mathbf{i}(3 \cdot 5 - 4 \cdot 4) \;-\; \mathbf{j}(2 \cdot 5 - 4 \cdot 3) \;+\; \mathbf{k}(2 \cdot 4 - 3 \cdot 3).
Evaluating:
= \mathbf{i}(15 - 16)\;-\;\mathbf{j}(10 - 12)\;+\;\mathbf{k}(8 - 9)
= -\mathbf{i} \;+\; 2\mathbf{j} \;-\; \mathbf{k}.
Thus,
\vec{d_1} \times \vec{d_2} = \langle -1,\; 2,\; -1 \rangle.
Its magnitude is
|\vec{d_1} \times \vec{d_2}| \;=\;\sqrt{(-1)^2 + 2^2 + (-1)^2} \;=\;\sqrt{1 + 4 + 1} \;=\;\sqrt{6}.
Step 5: Substitute into the distance formula and set the distance to 6
We know the shortest distance between these lines is given as 6 . Hence,
6 \;=\; \frac{\bigl|\bigl(\lambda + \sqrt{6},\; \sqrt{6},\; -3\sqrt{6}\bigr) \cdot \langle -1,\; 2,\; -1 \rangle\bigr|}{\sqrt{6}}.
Compute the dot product in the numerator:
(\lambda + \sqrt{6})(-1) \;+\; (\sqrt{6})(2) \;+\; (-3\sqrt{6})(-1)
= -(\lambda + \sqrt{6}) + 2\sqrt{6} + 3\sqrt{6}
= -\lambda - \sqrt{6} + 5\sqrt{6}
= -\lambda + 4\sqrt{6}.
Taking absolute value and dividing by \sqrt{6} must yield 6:
\frac{|-\lambda + 4\sqrt{6}|}{\sqrt{6}} \;=\; 6.
So
|-\lambda + 4\sqrt{6}| \;=\; 6 \sqrt{6}.
This leads to two possible equations:
1) -\lambda + 4\sqrt{6} = 6\sqrt{6} , or
2) -\lambda + 4\sqrt{6} = -6\sqrt{6}.
Step 6: Solve for the possible values of \lambda
From the first case:
-\lambda + 4\sqrt{6} = 6\sqrt{6}
\quad\Longrightarrow\quad
-\lambda = 6\sqrt{6} - 4\sqrt{6} = 2\sqrt{6}
\quad\Longrightarrow\quad
\lambda = -2\sqrt{6}.
From the second case:
-\lambda + 4\sqrt{6} = -6\sqrt{6}
\quad\Longrightarrow\quad
-\lambda = -6\sqrt{6} - 4\sqrt{6} = -10\sqrt{6}
\quad\Longrightarrow\quad
\lambda = 10\sqrt{6}.
Thus, the two possible values of \lambda are -2\sqrt{6} and 10\sqrt{6} .
Step 7: Find the square of the sum of all possible values
The sum of these values is:
(10\sqrt{6}) + (-2\sqrt{6}) = 8\sqrt{6}.
Hence, the square of this sum is:
\bigl(8\sqrt{6}\bigr)^2 = 64 \times 6 = 384.
Final Answer
The square of the sum of all possible values of \lambda is
\boxed{384}.
