Find the shortest distance between the lines
`vecr =lambda (2hati+ 3hatj+4hatk) and vecr=(hati-hatj)+t(2hati-3hatj+4hatk)`

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:
\[ \vec{r_1} = \lambda (2\hat{i} + 3\hat{j} + 4\hat{k}) \]
This can be interpreted as passing through the origin (0, 0, 0) with direction vector:
\[ \vec{b_1} = 2\hat{i} + 3\hat{j} + 4\hat{k} \]
So, we have:
\[ \vec{a_1} = (0, 0, 0) \]

The second line is given by:
\[ \vec{r_2} = (\hat{i} - \hat{j}) + t(2\hat{i} - 3\hat{j} + 4\hat{k}) \]
This can be interpreted as passing through the point (1, -1, 0) with direction vector:
\[ \vec{b_2} = 2\hat{i} - 3\hat{j} + 4\hat{k} \]
So, we have:
\[ \vec{a_2} = (1, -1, 0) \]

### Step 2: Check if the lines are parallel
To check if the lines are parallel, we need to see if the direction vectors \(\vec{b_1}\) and \(\vec{b_2}\) are proportional.

Calculating the cross product \(\vec{b_1} \times \vec{b_2}\):
\[ \vec{b_1} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}, \quad \vec{b_2} = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix} \]

Using the determinant to find the cross product:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 2 & -3 & 4 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i}(3 \cdot 4 - 4 \cdot (-3)) - \hat{j}(2 \cdot 4 - 4 \cdot 2) + \hat{k}(2 \cdot (-3) - 3 \cdot 2) \]
\[ = \hat{i}(12 + 12) - \hat{j}(8 - 8) + \hat{k}(-6 - 6) \]
\[ = 24\hat{i} + 0\hat{j} - 12\hat{k} = 24\hat{i} - 12\hat{k} \]

Since the cross product is not zero, the lines are not parallel.

### Step 3: Calculate the distance between the skew lines
The formula for the distance \(d\) between two skew lines is given by:
\[ d = \frac{|\vec{a_2} - \vec{a_1} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]

Calculating \(\vec{a_2} - \vec{a_1}\):
\[ \vec{a_2} - \vec{a_1} = (1, -1, 0) - (0, 0, 0) = (1, -1, 0) \]

Now, we need to find the dot product:
\[ \vec{a_2} - \vec{a_1} \cdot (\vec{b_1} \times \vec{b_2}) = (1, -1, 0) \cdot (24, 0, -12) \]
\[ = 1 \cdot 24 + (-1) \cdot 0 + 0 \cdot (-12) = 24 \]

Next, we calculate the magnitude of \(\vec{b_1} \times \vec{b_2}\):
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(24)^2 + (0)^2 + (-12)^2} = \sqrt{576 + 0 + 144} = \sqrt{720} = 12\sqrt{5} \]

### Step 4: Substitute values into the distance formula
Now substituting the values into the distance formula:
\[ d = \frac{|24|}{12\sqrt{5}} = \frac{24}{12\sqrt{5}} = \frac{2}{\sqrt{5}} \]

### Final Answer
Thus, the shortest distance between the two lines is:
\[ \boxed{\frac{2}{\sqrt{5}}} \]