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Step-by-Step Solution
Step 1: Identify Points and Direction Vectors of the Lines
The first line is given by
$ \frac{x - \lambda}{-2} = \frac{y - 2}{1} = \frac{z - 1}{1}. $
A convenient point on this line (for the parameter 0) is
$P_1(\lambda,\, 2,\, 1).$
Its direction vector is
$\vec{d_1} = (-2,\, 1,\, 1).$
The second line is given by
$ \frac{x - \sqrt{3}}{1} = \frac{y - 1}{-2} = \frac{z - 2}{1}. $
A convenient point on this line (for the parameter 0) is
$P_2(\sqrt{3},\, 1,\, 2).$
Its direction vector is
$\vec{d_2} = (1,\, -2,\, 1).$
Step 2: Express the Vector from One Point to the Other
The vector
$\vec{P_1P_2}$
from $P_1$ to $P_2$ is
$ (\sqrt{3} - \lambda,\; 1 - 2,\; 2 - 1). $
Thus,
$ \vec{P_1P_2} = (\sqrt{3} - \lambda,\; -1,\; 1). $
Step 3: Compute the Cross Product of the Direction Vectors
Next, calculate
$ \vec{d_1} \times \vec{d_2}: $
$
\vec{d_1} \times \vec{d_2}
=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-2 & 1 & 1 \\
1 & -2 & 1
\end{vmatrix}.
$
$
= \hat{i} \bigl(1 \cdot 1 - 1 \cdot (-2)\bigr)
\;-\;\hat{j} \bigl((-2)\cdot 1 - 1 \cdot 1\bigr)
\;+\;\hat{k} \bigl((-2)\cdot(-2) - 1 \cdot 1\bigr).
$
$
= (3)\hat{i} \;-\;\bigl(-3\bigr)\hat{j} \;+\;(4 - 1)\hat{k}
= (3,\; 3,\; 3).
$
Its magnitude is
$
|\vec{d_1} \times \vec{d_2}|
= \sqrt{3^2 + 3^2 + 3^2}
= \sqrt{27}
= 3\sqrt{3}.
$
Step 4: Use the 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 is given by
$
D
= \frac{\bigl|(\vec{P_1P_2}) \cdot (\vec{d_1} \times \vec{d_2})\bigr|}{|\vec{d_1} \times \vec{d_2}|}.
$
Step 5: Substitute the Values and Set the Distance to 1
First, compute
$
(\vec{P_1P_2}) \cdot (\vec{d_1} \times \vec{d_2}):
$
$
(\sqrt{3} - \lambda,\; -1,\; 1)
\cdot
(3,\; 3,\; 3)
= 3(\sqrt{3} - \lambda) + 3(-1) + 3(1).
$
$
= 3\sqrt{3} - 3\lambda - 3 + 3
= 3\sqrt{3} - 3\lambda.
$
Therefore,
$
D
= \frac{|3\sqrt{3} - 3\lambda|}{3\sqrt{3}}
= \frac{3|\sqrt{3} - \lambda|}{3\sqrt{3}}
= \frac{|\sqrt{3} - \lambda|}{\sqrt{3}}.
$
We are given $D = 1$, so
$
\frac{|\sqrt{3} - \lambda|}{\sqrt{3}} = 1
\;\;\Longrightarrow\;\;
|\sqrt{3} - \lambda| = \sqrt{3}.
$
Step 6: Solve for $ \lambda $
$
\sqrt{3} - \lambda = \pm \sqrt{3}.
$
• Case 1: $\sqrt{3} - \lambda = \sqrt{3} \;\Longrightarrow\; \lambda = 0.$
• Case 2: $\sqrt{3} - \lambda = -\sqrt{3} \;\Longrightarrow\; \lambda = 2\sqrt{3}.$
Step 7: Find the Sum of All Possible Values of $ \lambda $
The possible values of $\lambda$ are $0$ and $2\sqrt{3}.$
Their sum is
$
0 + 2\sqrt{3} = 2\sqrt{3}.
$
Final Answer: $2\sqrt{3}$
