The distance between the lines ⃗r = ˆ i− 2ˆ j + 3ˆ k + λ(2ˆ i+ 3ˆ j + 6ˆ k) and ⃗r = 3ˆ i − 2ˆ j + 1ˆ k + µ(4ˆ i+ 6ˆ j + 12ˆ k) is:

The correct option is √328/7.

Let's find the distance between the two given lines step-by-step.

The vector equations of the two lines are given by:
First line: $\overrightarrow{r}=(\hat{i}-2\hat{j}+3\hat{k})+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$
Second line: $\overrightarrow{r}=(3\hat{i}-2\hat{j}+\hat{k})+\mu (4\hat{i}+6\hat{j}+12\hat{k})$

Let us identify the position vectors of the points on the lines and their direction vectors:
For the first line:
Position vector, ${\overrightarrow{a}}_1=\hat{i}-2\hat{j}+3\hat{k}$
Direction vector, ${\overrightarrow{b}}_1=2\hat{i}+3\hat{j}+6\hat{k}$

For the second line:
Position vector, ${\overrightarrow{a}}_2=3\hat{i}-2\hat{j}+\hat{k}$
Direction vector, ${\overrightarrow{b}}_2=4\hat{i}+6\hat{j}+12\hat{k}=2(2\hat{i}+3\hat{j}+6\hat{k})=2{\overrightarrow{b}}_1$

Since the direction vector ${\overrightarrow{b}}_2$ is a scalar multiple of ${\overrightarrow{b}}_1$, the two lines are parallel to each other. We can write the common direction vector as:
$\overrightarrow{b}=2\hat{i}+3\hat{j}+6\hat{k}$

The shortest distance d between two parallel lines with equations $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda \stackrel{��}{b}$ and $\overrightarrow{r}={\overrightarrow{a}}_2+\mu \overrightarrow{b}$ is given by the formula:
$d=\frac{\left|\right({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)\times \overrightarrow{b}|}{\left|\overrightarrow{b}\right|}$

First, let's calculate the vector difference ${\overrightarrow{a}}_2-{\overrightarrow{a}}_1$:
${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=(3\hat{i}-2\hat{j}+\hat{k})-(\hat{i}-2\hat{j}+3\hat{k})$
${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=(3-1)\hat{i}+(-2-(-2\left)\right)\hat{j}+(1-3)\hat{k}$
${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=2\hat{i}+0\hat{j}-2\hat{k}$

Next, let's find the cross product $({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)\times \overrightarrow{b}$:
$({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 0& -2\\ 2& 3& 6\end{array}\right|$
$=\hat{i}\left[\right(0\left)\right(6)-(-2\left)\right(3\left)\right]-\hat{j}\left[\right(2\left)\right(6)-(-2\left)\right(2\left)\right]+\hat{k}\left[\right(2\left)\right(3)-(0\left)\right(2\left)\right]$
$=\hat{i}[0+6]-\hat{j}[12+4]+\hat{k}[6-0]$
$=6\hat{i}-16\hat{j}+6\hat{k}$

Now, let's find the magnitude of this cross product:
$\left|\right({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)\times \overrightarrow{b}|=\sqrt{6^2+{(-16)}^2+6^2}$
$=\sqrt{36+256+36}$
$=\sqrt{328}$

Next, let's find the magnitude of the direction vector $\overrightarrow{b}$:
$\left|\overrightarrow{b}\right|=\sqrt{2^2+3^2+6^2}$
$=\sqrt{4+9+36}$
$=\sqrt{49}=7$

Finally, we substitute these values into the distance formula:
$d=\frac{\sqrt{328}}{7}$

Thus, the distance between the two lines is $\frac{\sqrt{328}}{7}$.