The direction cosines x,y,z of a line are connected by a relation ${(x+y)}^2-a^2=2xy-z^2$ Then$\sqrt{a^2+5a+2019}=$

If $〈x,y,z〉$are direction cosines of a line then $x^2+y^2+z^2=1$

Consider the equation ${(x+y)}^2-a^2=2xy-z^2$

Expand and then simplify 

$\begin{array}{c}{x}^2+y^2+2xy-a^2=2xy-z^2\Rightarrow x^2+y^2+z^2=a^2\Rightarrow a^2=1\end{array}$

It implies that $a=1$

Hence the value of $\sqrt{a^2+5a+2019}$ is $\sqrt{1+5+2019}=\sqrt{2025}=\boxed{45}$