Find the shortest distance `vecr=hati+2hatj+3hatk+lambda(hati-3hatj+2hatk)and
vecr=
4hati+5hatj+6hatk+mu(2hati+3hatj+hatk)`.

Answer

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

### Step 1: Identify the lines
The two lines are given as:
1. \(\vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})\)
2. \(\vec{r_2} = 4\hat{i} + 5\hat{j} + 6\hat{k} + \mu(2\hat{i} + 3\hat{j} + \hat{k})\)

From these, we can identify:
- For line 1:
- \(\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}\)
- \(\vec{b_1} = \hat{i} - 3\hat{j} + 2\hat{k}\)

- For line 2:
- \(\vec{a_2} = 4\hat{i} + 5\hat{j} + 6\hat{k}\)
- \(\vec{b_2} = 2\hat{i} + 3\hat{j} + \hat{k}\)

### Step 2: Find \(\vec{a_2} - \vec{a_1}\)
Calculate the vector difference:
\[ \vec{a_2} - \vec{a_1} = (4\hat{i} + 5\hat{j} + 6\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (6 - 3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k} \]

### Step 3: Compute \(\vec{b_1} \times \vec{b_2}\)
Calculate the cross product \(\vec{b_1} \times \vec{b_2}\):
\[ \vec{b_1} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}, \quad \vec{b_2} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} \]
Using the determinant method:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}((-3)(1) - (2)(3)) - \hat{j}((1)(1) - (2)(2)) + \hat{k}((1)(3) - (-3)(2)) \]
\[ = \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 + 6) \]
\[ = -9\hat{i} + 3\hat{j} + 9\hat{k} \]

### Step 4: Find the magnitude of \(\vec{b_1} \times \vec{b_2}\)
Calculate the magnitude:
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19} \]

### Step 5: Calculate the shortest distance
Using the formula for the shortest distance \(d\) between two lines:
\[ d = \frac{|\vec{b_1} \times \vec{b_2} \cdot (\vec{a_2} - \vec{a_1})|}{|\vec{b_1} \times \vec{b_2}|} \]
Calculate the dot product:
\[ \vec{b_1} \times \vec{b_2} \cdot (\vec{a_2} - \vec{a_1}) = (-9\hat{i} + 3\hat{j} + 9\hat{k}) \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) \]
\[ = -9(3) + 3(3) + 9(3) = -27 + 9 + 27 = 9 \]
Thus, the distance is:
\[ d = \frac{|9|}{3\sqrt{19}} = \frac{9}{3\sqrt{19}} = \frac{3}{\sqrt{19}} \]

### Final Answer
The shortest distance between the two lines is:
\[ \frac{3}{\sqrt{19}} \]