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Step 1: Express Each Line in a Standard (Parametric) Form
The given lines are:
Line 1: 3(x − 1) = 6(y − 2) = 2(z − 1)
Line 2: 4(x − 2) = 2(y − λ) = (z − 3)
For Line 1, set 3(x − 1) = 6(y − 2) = 2(z − 1) = t (a parameter). Thus:
3(x − 1) = t ⟹ x − 1 = t/3 ⟹ x = 1 + t/3
6(y − 2) = t ⟹ y − 2 = t/6 ⟹ y = 2 + t/6
2(z − 1) = t ⟹ z − 1 = t/2 ⟹ z = 1 + t/2
So a point on Line 1 can be represented as
$ (x, y, z) = (1, 2, 1) + \left(\frac{t}{3}, \frac{t}{6}, \frac{t}{2}\right). $
The direction vector of Line 1 is
$ \vec{r}_1 = \left(\frac{1}{3}, \frac{1}{6}, \frac{1}{2}\right). $
For simplicity, we can multiply by 6 to get an equivalent direction vector
$ 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}. $
For Line 2, set 4(x − 2) = 2(y − λ) = (z − 3) = s (another parameter). Thus:
4(x − 2) = s ⟹ x − 2 = s/4 ⟹ x = 2 + s/4
2(y − λ) = s ⟹ y − λ = s/2 ⟹ y = λ + s/2
(z − 3) = s ⟹ z = 3 + s
So a point on Line 2 can be written as
$ (x, y, z) = (2, \lambda, 3) + \left(\frac{s}{4}, \frac{s}{2}, s\right). $
The direction vector of Line 2 is
$ \vec{r}_2 = \left(\frac{1}{4}, \frac{1}{2}, 1\right). $
For simplicity, multiplying by 4 gives an equivalent direction vector
$ \mathbf{i} + 2\mathbf{j} + 4\mathbf{k}. $
Step 2: Write the Vectors Clearly
Let a point on Line 1 be $ \vec{A} = (1, 2, 1) $ and a point on Line 2 be $ \vec{B} = (2, \lambda, 3). $
Then the direction vectors (scaled as convenient multiples) are:
$ \vec{r}_1 = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k} \quad \text{and} \quad \vec{r}_2 = \mathbf{i} + 2\mathbf{j} + 4\mathbf{k}. $
The vector $ \vec{a} = \overrightarrow{AB} = \vec{B} - \vec{A} = (2-1)\mathbf{i} + (\lambda-2)\mathbf{j} + (3-1)\mathbf{k} = \mathbf{i} + (\lambda - 2)\mathbf{j} + 2\mathbf{k}. $
Step 3: Use the Shortest Distance Formula Between Two Skew Lines
The shortest distance $D$ between two skew lines having direction vectors $ \vec{r}_1 $ and $ \vec{r}_2 $, and with a vector $ \vec{a} $ joining any point on one line to any point on the other, is given by:
$ D = \frac{\bigl|\vec{a}\cdot(\vec{r}_1 \times \vec{r}_2)\bigr|}{\bigl|\vec{r}_1 \times \vec{r}_2\bigr|}. $
We are given $D = \frac{1}{\sqrt{38}}.$
Step 4: Calculate the Cross Product $ \vec{r}_1 \times \vec{r}_2 $
$ \vec{r}_1 = (2, 1, 3), \quad \vec{r}_2 = (1, 2, 4). $
The cross product is:
$ \vec{r}_1 \times \vec{r}_2 =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
2 & 1 & 3 \\
1 & 2 & 4
\end{vmatrix}. $
Detailed expansion yields its magnitude $ \sqrt{38} $ (as also shown in the provided solution).
Step 5: Compute $ \vec{a} \cdot \bigl(\vec{r}_1 \times \vec{r}_2\bigr) $
From the provided determinant approach or direct vector methods,
$ \bigl|\vec{a} \cdot (\vec{r}_1 \times \vec{r}_2)\bigr| = |14 - 5\lambda| $
(this follows from the determinant involving the components of $ \vec{a}, \vec{r}_1, \vec{r}_2 $).
Thus,
$ D = \frac{|14 - 5\lambda|}{|\vec{r}_1 \times \vec{r}_2|} = \frac{|14 - 5\lambda|}{\sqrt{38}}. $
Given $ D = \frac{1}{\sqrt{38}}, $ we set:
$ \frac{|14 - 5\lambda|}{\sqrt{38}} = \frac{1}{\sqrt{38}}. $
Step 6: Solve for $ \lambda $
From the equation:
$ |14 - 5\lambda| = 1. $
This yields two cases:
1) $14 - 5\lambda = 1 \quad \Rightarrow \quad 5\lambda = 13 \quad \Rightarrow \quad \lambda = \frac{13}{5}.$
2) $14 - 5\lambda = -1 \quad \Rightarrow \quad 5\lambda = 15 \quad \Rightarrow \quad \lambda = 3.$
Step 7: Identify the Integral Value of $ \lambda $
The two possible values for $ \lambda $ are $ \frac{13}{5} $ and $ 3. $
The integral (integer) value requested is $ \lambda = 3. $
Therefore, the integral value of $ \lambda $ that satisfies the given shortest distance condition is
3.
