If

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

Now,

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

If $(\vec{a}+\lambda \vec{b})$ is perpendicular to $\vec{c}$, then

$(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0$

$\Rightarrow[(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k}] \cdot(3 \hat{i}+\hat{j})=0$

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

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

$\Rightarrow-\lambda+8=0$

$\Rightarrow \lambda=8$

Hence, the required value of $\lambda$ is 8 .