If the complex numbers $z_1,z_2,z_3$ and $z_4$ satisfy $5z_1-6z_2+3z_3-2z_4=0$ and $5{|z_1-z_4|}^2=9{|z_2-z_3|}^2$ then which of the following is/are correct

Let A,B,C,D denote the complex numbers $z_1,z_2,z_3\&z_4$ respectively in the argand plane , Now $\frac{5z_1+3z_3}{8}=\frac{6z_2+2z_4}{8}.$ P divides AC in the ration 3:5 & P divides BD in the ration 1:3

Let $AP=3\lambda ,CP=5\lambda , BP=k,DP=k.$ then in $\Delta APD$

$A{D}^2=A{P}^2+D{P}^2-2\left(AP\right)\left(DP\right)\cos \theta \Rightarrow {|z_1-z_4|}^2=9{\lambda }^2+9k^2-2\left(3\lambda \right)\left(3k\right)\cos \theta$

In $\Delta APD\Rightarrow {|z_2-z_3|}^2=25{\lambda }^2+k^2-2\left(5\lambda \right)\left(k\right)\cos \theta$ Given  

$5{|z_1-z_4|}^2=9{|z_2-z_3|}^2\Rightarrow 5\times 9({\lambda }^2+k^2)=9(25{\lambda }^2+k^2)\Rightarrow k^2=5{\lambda }^2$

$\therefore AP\times PC=BP\times PD\Rightarrow A,B,A,Dare\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}concyclic$