$\vec a = 2\hat i + 2\hat j + 3\hat k,; \vec b = -\hat i + 2\hat j + \hat k$ and $\vec c = 3\hat i + \hat j$ are such that $\vec a + \gamma \vec b$ is perpendicular to $\vec c$, then determine the value of $\gamma$.

Perpendicular condition: $(\vec a + \gamma \vec b)\cdot \vec c = 0$ $\vec a + \gamma \vec b = (2-\gamma)\hat i + (2+2\gamma)\hat j + (3+\gamma)\hat k$ $\vec c = 3\hat i + \hat j$ Dot product: $3(2-\gamma) + 1(2+2\gamma) = 0$ $6 - 3\gamma + 2 + 2\gamma = 0$ $8 - \gamma = 0 \Rightarrow \gamma = 8$