To find the shortest distance between the two given lines, we can use the formula for the distance between two skew lines in three-dimensional geometry. The lines are given in symmetric form, and we need to extract their direction ratios and points.
### Step-by-Step Solution:
1. **Identify the Lines**:
The lines are given as:
\[
L_1: \frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}
\]
\[
L_2: \frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}
\]
2. **Extract Points and Direction Ratios**:
For line \( L_1 \):
- Point \( A_1(3, -15, 9) \)
- Direction ratios \( \mathbf{b_1} = (2, -7, 5) \)
For line \( L_2 \):
- Point \( A_2(-1, 1, 9) \)
- Direction ratios \( \mathbf{b_2} = (2, 1, -3) \)
3. **Calculate the Vector from \( A_1 \) to \( A_2 \)**:
\[
\mathbf{A_2 - A_1} = (-1 - 3, 1 + 15, 9 - 9) = (-4, 16, 0)
\]
4. **Calculate the Cross Product \( \mathbf{b_1} \times \mathbf{b_2} \)**:
\[
\mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & -7 & 5 \\
2 & 1 & -3
\end{vmatrix}
\]
Expanding this determinant:
\[
= \mathbf{i}((-7)(-3) - (5)(1)) - \mathbf{j}((2)(-3) - (5)(2)) + \mathbf{k}((2)(1) - (-7)(2))
\]
\[
= \mathbf{i}(21 - 5) - \mathbf{j}(-6 - 10) + \mathbf{k}(2 + 14)
\]
\[
= 16\mathbf{i} + 16\mathbf{j} + 16\mathbf{k} = (16, 16, 16)
\]
5. **Calculate the Magnitude of the Cross Product**:
\[
|\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{16^2 + 16^2 + 16^2} = \sqrt{3 \times 16^2} = 16\sqrt{3}
\]
6. **Calculate the Dot Product**:
\[
(\mathbf{A_2 - A_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2}) = (-4, 16, 0) \cdot (16, 16, 16)
\]
\[
= -4 \times 16 + 16 \times 16 + 0 \times 16 = -64 + 256 = 192
\]
7. **Use the Distance Formula**:
The shortest distance \( d \) between the two lines is given by:
\[
d = \frac{|(\mathbf{A_2 - A_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|}
\]
\[
= \frac{|192|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}
\]
### Final Answer:
The shortest distance between the lines is \( 4\sqrt{3} \).
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