Let the angle between $u$ & $v$ be $\alpha$, now the dot product is given as follows $$u\cdot v=|u||v|\cos \alpha$$ $$\implies \cos \alpha=\frac{u\cdot v}{|u||v|}$$ $$=\frac{(1, -2, 3)\cdot (-4, 5, 6)}{|(1, -2, 3)||(-4, 5, 6)|}$$ $$\cos \alpha=\frac{-4-10+18}{\sqrt{(1)^2+(-2)^2+(3)^2}\sqrt{(-4)^2+(5)^2+(6)^2}}$$ $$\cos \alpha=\frac{4}{\sqrt{14\times 77}}$$$$\implies \color{blue}{\alpha=\cos^{-1}\left(\frac{4}{7\sqrt{22}}\right)\approx 83^\circ}$$
Cross product $|u\times v|$ is given as follows $$|u\times v|=\left|\begin{matrix} i&j&k\\ 1&-2&3\\-4&5&6
\end{matrix}\right|$$ $$=|-27i-18j-3k|$$ $$=\sqrt{(-27)^2+(-18)^2+(-3)^2}=\sqrt{1062}=3\sqrt{118}$$ Now, using cross product, the angle $\alpha$ between $u$ & $v$ is given as follows $$u\times v=|u||v|\sin \alpha(\hat n)$$ $$\implies |u\times v|=||u||v||\sin \alpha$$
$$\implies \sin\alpha=\frac{|u\times v|}{|u||v|}$$ $$=\frac{3\sqrt{118}}{7\sqrt{22}}=\frac{3}{7}\sqrt{\frac{59}{11}}$$
$$\color{blue}{\alpha=\sin^{-1}\left(\frac{3}{7}\sqrt{\frac{59}{11}}\right)\approx 83^\circ}$$
Your answer is correct