Let $$\vec{a}$$ be a vector which is perpendicular to the vector $$3\hat{i} + \frac{1}{2}\hat{j} + 2\hat{k}$$. If $$\vec{a} \times (2\hat{i} + \hat{k}) = 2\hat{i} - 13\hat{j} - 4\hat{k}$$, then the projection of the vector $$\vec{a}$$ on the vector $$2\hat{i} + 2\hat{j} + \hat{k}$$ is

Let $$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$. We are given two conditions.

Use the perpendicularity condition:

$$\vec{a} \cdot (3\hat{i} + \frac{1}{2}\hat{j} + 2\hat{k}) = 0$$

$$3a_1 + \frac{a_2}{2} + 2a_3 = 0 \quad \cdots (1)$$

Use the cross product condition:

$$\vec{a} \times (2\hat{i} + \hat{k}) = 2\hat{i} - 13\hat{j} - 4\hat{k}$$

$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 2 & 0 & 1 \end{vmatrix} = (a_2 - 0)\hat{i} - (a_1 - 2a_3)\hat{j} + (0 - 2a_2)\hat{k}$$

$$= a_2\hat{i} - (a_1 - 2a_3)\hat{j} - 2a_2\hat{k}$$

Comparing components:

$$a_2 = 2 \quad \cdots (2)$$

$$-(a_1 - 2a_3) = -13 \implies a_1 - 2a_3 = 13 \quad \cdots (3)$$

$$-2a_2 = -4 \implies a_2 = 2 \quad \checkmark$$

Solve the system:

From equation (1): $$3a_1 + 1 + 2a_3 = 0 \implies 3a_1 + 2a_3 = -1 \quad \cdots (4)$$

From equation (3): $$a_1 = 13 + 2a_3$$

Substituting into (4): $$3(13 + 2a_3) + 2a_3 = -1$$

$$39 + 6a_3 + 2a_3 = -1$$

$$8a_3 = -40 \implies a_3 = -5$$

$$a_1 = 13 + 2(-5) = 3$$

So $$\vec{a} = 3\hat{i} + 2\hat{j} - 5\hat{k}$$.

Find the projection:

Projection of $$\vec{a}$$ on $$\vec{c} = 2\hat{i} + 2\hat{j} + \hat{k}$$:

$$\text{proj} = \frac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = \frac{6 + 4 - 5}{\sqrt{4 + 4 + 1}} = \frac{5}{3}$$

The correct answer is Option C: $$\boxed{\dfrac{5}{3}}$$.