Find the shortest distacne between the lines
`vecr=hati+2hatj+3hatk+mu(2hati+3hatj+4hatk)` and `vecr=(2hati+4hatj+5hatk)+lambda(3hati+4hatj+5hatk)`

Answer

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

### Step 1: Identify the lines and their components
The first line is given by:
\[ \vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k} + \mu(2\hat{i} + 3\hat{j} + 4\hat{k}) \]
From this, we can identify:
- Point \( \vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k} \)
- Direction vector \( \vec{b_1} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)

The second line is given by:
\[ \vec{r_2} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \lambda(3\hat{i} + 4\hat{j} + 5\hat{k}) \]
From this, we can identify:
- Point \( \vec{a_2} = 2\hat{i} + 4\hat{j} + 5\hat{k} \)
- Direction vector \( \vec{b_2} = 3\hat{i} + 4\hat{j} + 5\hat{k} \)

### Step 2: Calculate the cross product of the direction vectors
To find the shortest distance, we need to calculate the cross product \( \vec{b_1} \times \vec{b_2} \).

\[ \vec{b_1} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}, \quad \vec{b_2} = \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix} \]

The cross product is calculated using the determinant:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \left( 3 \cdot 5 - 4 \cdot 4 \right) - \hat{j} \left( 2 \cdot 5 - 4 \cdot 3 \right) + \hat{k} \left( 2 \cdot 4 - 3 \cdot 3 \right) \]
\[ = \hat{i} (15 - 16) - \hat{j} (10 - 12) + \hat{k} (8 - 9) \]
\[ = -\hat{i} + 2\hat{j} - \hat{k} \]

### Step 3: Calculate \( \vec{a_1} - \vec{a_2} \)
Now we calculate the vector from point \( \vec{a_2} \) to point \( \vec{a_1} \):
\[ \vec{a_1} - \vec{a_2} = (\hat{i} + 2\hat{j} + 3\hat{k}) - (2\hat{i} + 4\hat{j} + 5\hat{k}) \]
\[ = (-1)\hat{i} + (-2)\hat{j} + (-2)\hat{k} \]

### Step 4: Calculate the magnitude of the cross product
The magnitude of the cross product \( |\vec{b_1} \times \vec{b_2}| \) is:
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

### Step 5: Calculate the magnitude of \( \vec{a_1} - \vec{a_2} \)
The magnitude of \( \vec{a_1} - \vec{a_2} \) is:
\[ |\vec{a_1} - \vec{a_2}| = \sqrt{(-1)^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Step 6: Calculate the shortest distance
The formula for the shortest distance \( d \) between two skew lines is:
\[ d = \frac{|\vec{b_1} \times \vec{b_2} \cdot (\vec{a_1} - \vec{a_2})|}{|\vec{b_1} \times \vec{b_2}|} \]

Calculating the dot product:
\[ \vec{b_1} \times \vec{b_2} \cdot (\vec{a_1} - \vec{a_2}) = (-1)\cdot(-1) + 2\cdot(-2) + (-1)\cdot(-2) = 1 - 4 + 2 = -1 \]

Thus, the shortest distance is:
\[ d = \frac{|-1|}{\sqrt{6}} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6} \]

### Final Answer
The shortest distance between the two lines is:
\[ \frac{\sqrt{6}}{6} \text{ units} \]