The magnitude of the projection of the vector $$2\hat{i} + 3\hat{j} + \hat{k}$$ on the vector perpendicular to the plane containing the vectors $$\hat{i} + \hat{j} + \hat{k}$$ and $$\hat{i} + 2\hat{j} + 3\hat{k}$$, is:

We have the vector whose projection is required, $$\vec{A}=2\hat{i}+3\hat{j}+\hat{k}.$$ The projection is to be taken on the vector that is perpendicular to the plane containing the two given vectors $$\vec{B}=\hat{i}+\hat{j}+\hat{k} \text{ and } \vec{C}=\hat{i}+2\hat{j}+3\hat{k}.$$

A vector perpendicular to a plane containing two non-parallel vectors is obtained through their cross product. Stating the formula first, for any vectors $$\vec{u}=(u_1,u_2,u_3)$$ and $$\vec{v}=(v_1,v_2,v_3),$$ the cross product is

$$\vec{u}\times\vec{v}= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ u_1&u_2&u_3\\ v_1&v_2&v_3 \end{vmatrix}.$$

Now we compute $$\vec{N}=\vec{B}\times\vec{C}:$$

$$ \vec{N}= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 1&1&1\\ 1&2&3 \end{vmatrix} = \hat{i}(1\cdot3-1\cdot2) - \hat{j}(1\cdot3-1\cdot1) + \hat{k}(1\cdot2-1\cdot1). $$

Simplifying each component, we get

$$ \vec{N}= \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1)= \hat{i} - 2\hat{j} + \hat{k}. $$

So the perpendicular (normal) vector is $$\vec{N}= \hat{i}-2\hat{j}+\hat{k}.$$

To find the magnitude of the projection of $$\vec{A}$$ on $$\vec{N},$$ we state the projection magnitude formula: for vectors $$\vec{u}$$ and $$\vec{v},$$ the magnitude of the projection of $$\vec{u}$$ on $$\vec{v}$$ is

$$ \left|\text{proj}_{\vec{v}}\vec{u}\right| = \frac{|\vec{u}\cdot\vec{v}|}{\|\vec{v}\|}. $$

We therefore need $$\vec{A}\cdot\vec{N}$$ and $$\|\vec{N}\|.$$

First, the dot product:

$$ \vec{A}\cdot\vec{N} = (2)(1) + (3)(-2) + (1)(1) = 2 - 6 + 1 = -3. $$

The absolute value is $$|\vec{A}\cdot\vec{N}| = |-3| = 3.$$

Next, the magnitude of $$\vec{N}$$ is

$$ \|\vec{N}\| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}. $$

Substituting these results in the projection formula gives

$$ \left|\text{proj}_{\vec{N}}\vec{A}\right| = \frac{3}{\sqrt{6}}. $$

We rationalise the denominator to make comparison with the options easier:

$$ \frac{3}{\sqrt{6}} = \frac{3}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2}. $$

An equivalent way to write $$\frac{3}{\sqrt{6}}$$ is $$\sqrt{\frac{3}{2}}$$ because

$$ \sqrt{\frac{3}{2}}=\sqrt{\frac{9}{6}}=\frac{3}{\sqrt{6}}. $$

Among the given choices, this matches option B, which is $$\sqrt{\frac{3}{2}}.$$

Hence, the correct answer is Option 2.