Find the shortest distance between the lines
`(x)/2=(y-2)/3=(z-4)/3 and (x-1)/3=(y-2)/7=(z-3)/8`

Answer

To find the shortest distance between the two lines given by the equations

1. \(\frac{x}{2} = \frac{y-2}{3} = \frac{z-4}{3}\)
2. \(\frac{x-1}{3} = \frac{y-2}{7} = \frac{z-3}{8}\)

we will follow these steps:

### Step 1: Write the parametric equations of the lines

For the first line, we can express it in vector form as:
\[ \mathbf{r_1} = \begin{pmatrix} 0 \\ 2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 3 \\ 3 \end{pmatrix} \]
where \(\lambda\) is a parameter.

For the second line, we can express it as:
\[ \mathbf{r_2} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 7 \\ 8 \end{pmatrix} \]
where \(\mu\) is another parameter.

### Step 2: Identify direction vectors and points on the lines

From the equations, we identify:
- For Line 1:
- Point \(A_1 = (0, 2, 4)\)
- Direction vector \(\mathbf{b_1} = \begin{pmatrix} 2 \\ 3 \\ 3 \end{pmatrix}\)

- For Line 2:
- Point \(A_2 = (1, 2, 3)\)
- Direction vector \(\mathbf{b_2} = \begin{pmatrix} 3 \\ 7 \\ 8 \end{pmatrix}\)

### Step 3: Compute the cross product of the direction vectors

We need to compute \(\mathbf{b_1} \times \mathbf{b_2}\):
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 3 \\ 3 & 7 & 8 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i}(3 \cdot 8 - 3 \cdot 7) - \mathbf{j}(2 \cdot 8 - 3 \cdot 3) + \mathbf{k}(2 \cdot 7 - 3 \cdot 3) \]
\[ = \mathbf{i}(24 - 21) - \mathbf{j}(16 - 9) + \mathbf{k}(14 - 9) \]
\[ = 3\mathbf{i} - 7\mathbf{j} + 5\mathbf{k} \]

### Step 4: Find the magnitude of the cross product

The magnitude of \(\mathbf{b_1} \times \mathbf{b_2}\) is:
\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{3^2 + (-7)^2 + 5^2} = \sqrt{9 + 49 + 25} = \sqrt{83} \]

### Step 5: Compute the vector \(A_2 - A_1\)

Now, we compute the vector from point \(A_1\) to point \(A_2\):
\[ \mathbf{A_2 - A_1} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} - \begin{pmatrix} 0 \\ 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \]

### Step 6: Compute the dot product

Next, we compute the dot product:
\[ (\mathbf{A_2 - A_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2}) = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -7 \\ 5 \end{pmatrix} = 1 \cdot 3 + 0 \cdot (-7) + (-1) \cdot 5 = 3 - 5 = -2 \]

### Step 7: Calculate the shortest distance

Finally, we can use the formula for the shortest distance \(d\) between the two lines:
\[ d = \frac{|(\mathbf{A_2 - A_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]
Substituting the values we found:
\[ d = \frac{|-2|}{\sqrt{83}} = \frac{2}{\sqrt{83}} \]

### Final Answer
Thus, the shortest distance between the two lines is:
\[ \frac{2}{\sqrt{83}} \text{ units} \]