Let $$\vec{a} = \hat{i} + \hat{j} + 2\hat{k}, \vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$$ and $$\vec{c} = \hat{i} - \hat{j} + \hat{k}$$ be the three given vectors. Let $$\vec{v}$$ be a vector in the plane of $$\vec{a}$$ and $$\vec{b}$$ whose projection on $$\vec{c}$$ is $$\frac{2}{\sqrt{3}}$$. If $$\vec{v} \cdot \hat{j} = 7$$, then $$\vec{v} \cdot (\hat{i} + \hat{k})$$ is equal to

Given $$\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$$, $$\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$$, and $$\vec{c} = \hat{i} - \hat{j} + \hat{k}$$.

First, we express $$\vec{v}$$ in the plane of $$\vec{a}$$ and $$\vec{b}$$ as $$\vec{v} = \lambda\vec{a} + \mu\vec{b} = (\lambda + 2\mu)\hat{i} + (\lambda - 3\mu)\hat{j} + (2\lambda + \mu)\hat{k}$$.

Next, using the condition $$\vec{v} \cdot \hat{j} = 7$$ yields $$\lambda - 3\mu = 7 \quad \cdots (1)$$.

Since the projection of $$\vec{v}$$ on $$\vec{c}$$ is $$\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = \frac{2}{\sqrt{3}},$$ we note that $$|\vec{c}| = \sqrt{1+1+1} = \sqrt{3}$$ and

$$\vec{v} \cdot \vec{c} = (\lambda + 2\mu)(1) + (\lambda - 3\mu)(-1) + (2\lambda + \mu)(1) = \lambda + 2\mu - \lambda + 3\mu + 2\lambda + \mu = 2\lambda + 6\mu.$$

Therefore, $$\frac{2\lambda + 6\mu}{\sqrt{3}} = \frac{2}{\sqrt{3}} \implies 2\lambda + 6\mu = 2 \implies \lambda + 3\mu = 1 \quad \cdots (2).$$

Now, from (1) we have $$\lambda = 7 + 3\mu$$, and substituting into (2) gives $$7 + 3\mu + 3\mu = 1 \implies 6\mu = -6 \implies \mu = -1$$ and hence $$\lambda = 7 + 3(-1) = 4$$.

Substituting these into the expression for $$\vec{v}$$ yields $$\vec{v} = (4 + 2(-1))\hat{i} + (4 - 3(-1))\hat{j} + (8 + (-1))\hat{k} = 2\hat{i} + 7\hat{j} + 7\hat{k}$$.

Finally, computing the required dot product gives $$\vec{v} \cdot (\hat{i} + \hat{k}) = 2 + 7 = 9$$.

Hence the correct answer is Option D: $$9$$.