$\left(a^{2}+b^{2}+c^{2}\right) p^{2}+2(a b+b c+c d) p+b^{2}+c^{2}+d^{2}$

$=0$

$\Rightarrow\left(a^{2} p^{2}+2 a b p+b^{2}\right)+\left(b^{2} p^{2}+2 b c p+c^{2}\right)+$

$\left(c^{2} p^{2}+2 c d p+d^{2}\right)=0$

$\Rightarrow(a b+b)^{2}+(b p+c)^{2}+(c p+d)^{2}=0$

This is possible only when $a p+b=0$ and $b p+c=0$ and $c p+d=0$

$p =-\frac{ b }{ a }=-\frac{ c }{ b }=-\frac{ d }{ c }$

or $\frac{ b }{ a }=\frac{ c }{ b }=\frac{ d }{ c }$

$\therefore a , b , c , d$ are in $G . P$