Given $3$ $\sin \,P + 4\cos Q = 6$ $\,\,\,\,\,\,\,\,...\left( i \right)$ $4\sin Q + 3\cos P = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$ Squaring and adding $(i)$ & $(ii)$ we get $9\,{\sin ^2}P + 16{\cos ^2}Q + 24\sin P\cos Q$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 16\,{\sin ^2}Q + 9{\cos ^2}P + 24\sin Q\cos P$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $= 36 + 1 = 37$ $\Rightarrow 9\left( {{{\sin }^2}p + {{\cos }^2}P} \right) + 16\left( {{{\sin }^2}Q + {{\cos }^2}q} \right)$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 24\left( {\sin P\cos Q + \cos P\sin Q} \right) = 37$ $\Rightarrow 9 + 16 + 24\sin \left( {P + Q} \right) = 37$ [ As ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and $\sin A\cos B + \cos A\sin B$ $= \sin \left( {A + B} \right)$ ] $\Rightarrow \sin \left( {P + Q} \right) = {1 \over 2}$ $\Rightarrow P + Q = {\pi \over 6}$ or ${{5\pi } \over 6}$ $\Rightarrow R = {{5\pi } \over 6}$ or ${\pi \over 6}$ (as $P + Q + R = \pi$ ) If $R = {{5\pi } \over 6}$ then $0 < P,Q < {\pi \over 6}$ $\Rightarrow \cos Q < 1$ and $\sin P < {1 \over 2}$ $\Rightarrow 3\sin P + 4\cos Q < {{11} \over 2}$ which is not true. So $R = {\pi \over 6}$