the resultant of two vectors 3Pand2P is R. if first is doubled the fresultant also gets doubled. then find the angle between two forces

$$\begin{gathered} Lettheanglebetween3Pand2Pvectorsis\theta ,sotheresul\tan tRisgivenby, \\ {R}^2={\left(3P\right)}^2+{\left(2P\right)}^2+2\times 3P\times 2P\times \cos \theta . \\ {R}^2=9P^2+4P^2+12P^2\cos \theta =13P^2+12P^2\cos \theta ---\left(1\right). \\ Whenfirstvectorisdoubledi.e.made6P,thenresul\tan tbecomes2R.Therefore \\ {\left(2R\right)}^2={\left(6P\right)}^2+{\left(2P\right)}^2+2\times 6P\times 2P\times \cos \theta .OR \\ 4R^2=36P^2+4P^2+24P^2\cos \theta =40P^2+24P^2\cos \theta ----\left(2\right) \\ Puttingthevalueof{R}^{2}fromequation\left(1\right)toeqation\left(2\right). \\ 4\left(13P^2+12P^2\cos \theta \right)=40P^2+24P^2\cos \theta .or52P^2+36P^2\cos \theta =40P^2+24P^2\cos \theta . \\ 12P^2=-12P^2\cos \theta or\cos \theta =-1or\theta =\pi . \\ Thustheanglebetweenthetwovectosis\pi =180°. \end{gathered}$$