Given $\vec{A}=(2 \hat{i}-3 \hat{j}+\hat{k}), \vec{B}=(3 \hat{i}+\hat{j}-2 \hat{k})$ and $\vec{C}=(3 \hat{i}+2 \hat{j}+\hat{k})$. $(\vec{A}+\vec{B}) \cdot \vec{C}$ will be

Given vectors: $\vec{A} = (2\hat{i}-3\hat{j}+\hat{k})$, $\vec{B} = (3\hat{i}+\hat{j}-2\hat{k})$, $\vec{C} = (3\hat{i}+2\hat{j}+\hat{k})$

The vector sum $\vec{A} + \vec{B}$ is found by adding corresponding components:

$\vec{A} + \vec{B} = (2+3)\hat{i} + (-3+1)\hat{j} + (1-2)\hat{k} = 5\hat{i} - 2\hat{j} - \hat{k}$

The dot product with $\vec{C}$ is computed as:

$(\vec{A}+\vec{B}) \cdot \vec{C} = (5)(3) + (-2)(2) + (-1)(1) = 15 - 4 - 1 = 10$

The result is $\boxed{10}$.