Given, $\overrightarrow{a}\times \left(2\hat{i}+\hat{k}\right)=2\hat{i}-13\hat{j}-4\hat{k}$

Now multiplying both side by $\left(3\hat{i}+\frac{1}{2}\hat{j}+2\hat{k}\right)$ we get,

$\Rightarrow \left(\overrightarrow{a}\times \left(2\hat{i}+\hat{k}\right)\right)\times \left(3\hat{i}+\frac{1}{2}\hat{j}+2\hat{k}\right)$

$=\left(2\hat{i}-13\hat{j}-4\hat{k}\right)\times \left(3\hat{i}+\frac{1}{2}\hat{j}+2\hat{k}\right)$

$\Rightarrow \left(\left(3\hat{i}+\frac{1}{2}\hat{j}+2\hat{k}\right).\overrightarrow{a}\right)\left(2\hat{i}+\hat{k}\right)-\left(6+2\right)\overrightarrow{a}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -13& -4\\ 3& \frac{1}{2}& 2\end{array}\right|$

$\Rightarrow 0-\left(6+2\right)\overrightarrow{a}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -13& -4\\ 3& \frac{1}{2}& 2\end{array}\right|$

$\overrightarrow{a}=3\hat{i}+2\hat{j}-5\hat{k}$

Now, projection of $\overrightarrow{a}$ on vector $2\hat{i}+2\hat{j}+\hat{k}$ is $\overrightarrow{a}·\frac{\left(2\hat{i}+2\hat{j}+\hat{k}\right)}{3}=\frac{5}{3}$