Let $\vec{u}=\hat{j}+4 \hat{k}, \vec{v}=\hat{i}-3 \hat{k}$ and
$$
\vec{w}=\cos \theta \hat{i}+\sin \theta \hat{j}
$$
Now, $\vec{u} \times \vec{v}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 4 \\ 1 & 0 & -3\end{array}\right|$
$$
=\hat{i}(-3+\hat{j}-4(+\hat{k})-1 \quad(\quad)
$$
$$
=-3 \hat{i}+4 \hat{j}-\hat{k}
$$
Now,
$$
\begin{aligned}
& (\vec{u} \times \vec{v}) \cdot \vec{w}=(-3 \hat{i}+4 \hat{j}-\hat{k}) \cdot(\cos \theta \hat{i}+\sin \theta \hat{j}) \\
& =-3 \cos \theta+4 \sin \theta
\end{aligned}
$$
Now, maximum possible value of
$$
|-3 \cos \theta+4 \sin \theta|=\sqrt{(-3)^2+(4)^2}=\sqrt{25}=5
$$