Let a,b are roots of the equation $x^2-10cx-11d=0$ and c,d are the roots of $x^2-10ax-11b=0$. If $a+b+c+d=p^{2}q$ where $p$ is prime number then the value of $p-q\left(a\ne b\ne c\ne d\right)$ is

$a+b=10c,ab=-11d$

$c+d=10a,cd=-11b$

$\Rightarrow a+b+c+d=10(a+c),abcd=121bd$

$\Rightarrow b+d=9(a+c),ac=121$

and also $a^2-10ac-11d=0,c^2-10ac-11b=0$

$\Rightarrow a^2+c^2-20ac-11(b+d)=0$

$\Rightarrow (a+c{)}^2-22ac-99(a+c)=0$

$\Rightarrow a+c=-121\left(\text{or}\right)-22$

If $a+c=-22 \text{then }a=c$

$\therefore a+c=121\Rightarrow a+b+c+d=1210$

$=(11{)}^2\times 10$