$\text{ A triangle }\mathrm{ABC}\text{ satisfies the relation }2\sec 4\mathrm{C}+{\sin }^{2}2 \mathrm{A}+$$\sqrt{\mathrm{sinB}}=0\text{ and a point }\mathrm{P}\text{ is taken on the longest side of the }$$\text{ triangle such that it divides the side in the ratio }1:3\text{ . Let }\mathrm{Q}$$\text{ and }\mathrm{R}\text{ be the circumcentre and orthocentre of }\Delta \mathrm{ABC}\text{ . IfPQ : }$$\mathrm{QR}:\mathrm{RP}=1:\alpha :\beta ,\text{ then the value of }{\alpha }^2+{\beta }^2.$

$\begin{array}{l}\text{Given }2sec4C+si{n}^{2}2A+\sqrt{\sin B}=0\\ \text{This is possible when the triangle ABC }\\ \text{is right angled Isosceles triangle with right angle at }B\\ \text{Since R is orthocenter, it is coincide with B}\\ \text{Since Q is circum center it is midpoint of AB}\\ \text{Suppose that QR}=2x,AQ=2x\\ \text{Since the point }P\text{\hspace{0.17em} divides the hypotenuse in the ration 1:3}\\ \text{Hence, }PQ=x\\ \text{The triangle PQR\hspace{0.17em}is right angled triangle}\\ \text{Hence, }\left(\text{PR}\right)=\sqrt{{\left(PQ\right)}^2+{\left(QR\right)}^2}=\sqrt{4x^2+x^2}=\sqrt{5}x\\ \text{Therefore, PQ:QR:RP}=1:2:\sqrt{5}\\ \text{hence, }{\alpha }^2+{\beta }^2=4+5=9\end{array}$