The shortest distance between the line `(x-2)/3=(y+1)/2=(z-6)/2` and `(x-6)/3=(1-y)/2=(z+8)/0` is equal to`

Answer

To find the shortest distance between the two given lines, we can follow these steps:

### Step 1: Identify the lines and their parametric equations
The first line can be expressed in parametric form as:
\[ L_1: \begin{cases} x = 2 + 3t \\ y = -1 + 2t \\ z = 6 + 2t \end{cases} \]
where \( t \) is a parameter.

The second line can be expressed as:
\[ L_2: \begin{cases} x = 6 + 3s \\ y = 1 - 2s \\ z = -8 \end{cases} \]
where \( s \) is another parameter.

### Step 2: Identify points and direction vectors
From the parametric equations, we can identify:
- A point on line \( L_1 \) is \( A(2, -1, 6) \) and the direction vector \( \mathbf{p} = \langle 3, 2, 2 \rangle \).
- A point on line \( L_2 \) is \( B(6, 1, -8) \) and the direction vector \( \mathbf{q} = \langle 3, -2, 0 \rangle \).

### Step 3: Find the vector \( \mathbf{AB} \)
The vector \( \mathbf{AB} \) from point \( A \) to point \( B \) is:
\[ \mathbf{AB} = \mathbf{B} - \mathbf{A} = \langle 6 - 2, 1 - (-1), -8 - 6 \rangle = \langle 4, 2, -14 \rangle \]

### Step 4: Calculate the cross product \( \mathbf{p} \times \mathbf{q} \)
To find the shortest distance, we need the cross product of the direction vectors \( \mathbf{p} \) and \( \mathbf{q} \):
\[ \mathbf{p} \times \mathbf{q} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & 2 \\ 3 & -2 & 0 \end{vmatrix} \]
Calculating this determinant:
\[ = \mathbf{i} \left(2 \cdot 0 - 2 \cdot (-2)\right) - \mathbf{j} \left(3 \cdot 0 - 2 \cdot 3\right) + \mathbf{k} \left(3 \cdot (-2) - 2 \cdot 3\right) \]
\[ = \mathbf{i}(0 + 4) - \mathbf{j}(0 - 6) + \mathbf{k}(-6 - 6) \]
\[ = 4\mathbf{i} + 6\mathbf{j} - 12\mathbf{k} \]
Thus, \( \mathbf{p} \times \mathbf{q} = \langle 4, 6, -12 \rangle \).

### Step 5: Calculate the magnitude of the cross product
The magnitude of \( \mathbf{p} \times \mathbf{q} \) is:
\[ |\mathbf{p} \times \mathbf{q}| = \sqrt{4^2 + 6^2 + (-12)^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14 \]

### Step 6: Calculate the shortest distance
The shortest distance \( D \) between the two lines is given by:
\[ D = \frac{|\mathbf{AB} \cdot (\mathbf{p} \times \mathbf{q})|}{|\mathbf{p} \times \mathbf{q}|} \]
Calculating \( \mathbf{AB} \cdot (\mathbf{p} \times \mathbf{q}) \):
\[ \mathbf{AB} \cdot (\mathbf{p} \times \mathbf{q}) = \langle 4, 2, -14 \rangle \cdot \langle 4, 6, -12 \rangle = 4 \cdot 4 + 2 \cdot 6 + (-14) \cdot (-12) \]
\[ = 16 + 12 + 168 = 196 \]
Thus,
\[ D = \frac{|196|}{14} = \frac{196}{14} = 14 \]

### Final Answer
The shortest distance between the two lines is \( \boxed{14} \).

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