$\dfrac{1}{\cos^2\alpha}=1+\tan^2\alpha$ $\to \cos^2\alpha=\dfrac{4}{13}$
$\cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{2}{3}$
$\to Q=3\sin^2\alpha+5\cos^2\alpha+\cot\alpha=3(\sin^2\alpha+\cos^2\alpha)+2\cos^2\alpha+\cot\alpha$
$=3.1+2.\dfrac{4}{13}+\dfrac{2}{3}$
$=\dfrac{167}{39}$