In a  $\Delta PQR,$  if  $3\sin P+4\cos Q=6$  and  $4\sin Q+3\cos P=1,$ then the angle R equal to 

$3\sin P+4\cos Q=\mathrm{6.............}\left(i\right)$

$\Rightarrow 4\sin Q+3\cos P=\mathrm{1.............}\left(ii\right)$

Squaring and adding (i) and (ii)
$$\begin{gathered} 9{\sin }^{2}P+16{\cos }^{2}Q+24\sin P\sin Q+16{\sin }^{2}Q+9co{s}^{2}P+24\cos P\sin Q=37 \\ \Rightarrow 9+16+24(sinP\cos Q+\cos P\sin Q)=37 \\ \Rightarrow 24\sin (P+Q)=12 \end{gathered}$$ 

$\Rightarrow \sin \left(P+Q\right)=\frac{1}{2}\Rightarrow P+Q=\frac{\pi }{6}or\frac{5\pi }{6}$

$\Rightarrow 3\sin P+4\cos Q<\frac{11}{2},$ which  is not possible; 

 So  $R=\frac{\pi }{6}$