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

Here, $\vec a = 2\hat i + 2\hat j + 3\hat k,\,\,\vec b = - \hat i + 2\hat j + \hat k$ and $\vec c = 3\hat i + \hat j$

Now, $\vec a + \lambda \vec b = (2\hat i + 2\hat j + 3\hat k) + \lambda \left( { - \hat i + 2\hat j + \hat k} \right)$

$= \left( {2 - \lambda } \right)\hat i + \left( {2 + 2\lambda } \right)\hat j + \left( {3 + \lambda } \right)\hat k$

Since, $\left( {a + \lambda b} \right)$ is perpendicular to $\vec c\,\,\,\therefore \,\,(\vec a + \lambda \vec b) \cdot \vec c = 0$

$\Rightarrow (2 - \lambda )(3) + (2 + 2\lambda )(1) + (3 + \lambda )(0) = 0$

$\Rightarrow 6 - 3\lambda + 2 + 2\lambda = 0 \Rightarrow \lambda = 8$