Two bodies of mass 1 kg and 3 kg have position vectors $$\hat{i} + 2\hat{j} + \hat{k}$$ and $$-3\hat{i} - 2\hat{j} + \hat{k}$$ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector:

We have two bodies of masses $$m_1 = 1$$ kg and $$m_2 = 3$$ kg with position vectors $$\vec{r}_1 = \hat{i} + 2\hat{j} + \hat{k}$$ and $$\vec{r}_2 = -3\hat{i} - 2\hat{j} + \hat{k}$$ respectively.

The position vector of the centre of mass is $$\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3}$$.

Computing the numerator: $$(1 - 9)\hat{i} + (2 - 6)\hat{j} + (1 + 3)\hat{k} = -8\hat{i} - 4\hat{j} + 4\hat{k}$$.

Dividing by 4: $$\vec{r}_{cm} = -2\hat{i} - \hat{j} + \hat{k}$$.

The magnitude is $$|\vec{r}_{cm}| = \sqrt{(-2)^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$.

Now we check the magnitudes of each option. For Option A: $$\hat{i} - 2\hat{j} + \hat{k}$$, the magnitude is $$\sqrt{1 + 4 + 1} = \sqrt{6}$$. This matches our answer.

Hence, the correct answer is Option A.