Let, $C=\left(x,y,z\right)$ such that $\overrightarrow{BC}$ is parallel to $\overrightarrow{OA}\times \overrightarrow{OB}$ and $\left|\overrightarrow{BC}\right|=\sqrt{189}$.

Apply $\overrightarrow{BC}=\lambda \overrightarrow{OA}\times \overrightarrow{OB}$

$\Rightarrow \left(x-4\right)\hat{i}+y\hat{j}+\left(z-1\right)\hat{k}=\lambda \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 4& 0& 1\end{array}\right|$

$\Rightarrow \left(x-4\right)\hat{i}+y\hat{j}+\left(z-1\right)\hat{k}=\lambda \left[2\hat{i}+11\hat{j}-8\hat{k}\right]$

$\Rightarrow x=4+2\lambda , y=11\lambda , z=1-8\lambda$

Apply $\left|\overrightarrow{BC}\right|=\sqrt{189}$

$\Rightarrow \sqrt{{\left(2\lambda \right)}^2+{\left(11\lambda \right)}^2+{\left(-8\lambda \right)}^2}=\sqrt{189}$

$\Rightarrow \lambda =1$

So the position vector of the point $C$ is $6\hat{i}+11\hat{j}-7\hat{k}$.