The circum centre of the triangle whose sides are $3\mathrm{x}-\mathrm{y}-5=0,\mathrm{x}+2\mathrm{y}-4=0$,$5\mathrm{x}+3\mathrm{y}+1=0$  is $\left(\mathrm{\alpha },\mathrm{\beta }\right)$ then  ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=$

Given lines are

  $3\mathrm{x}-\mathrm{y}-5=0$       …… (1)
 $\mathrm{x}+2\mathrm{y}-4=0$       …… (2)
   $5\mathrm{x}+3\mathrm{y}+1=0$     …… (3)
The point of intersections of above lines are vertices of triangle 
Hence the vertices are   $\mathrm{A}\left(2,1\right),\mathrm{B}\left(-2,3\right),\mathrm{C}\left(1,-2\right)$
Suppose that the point $\mathrm{S}\left(\mathrm{\alpha },\mathrm{\beta }\right)$ is the circum center of the triangle, so that  $\mathrm{SA}=\mathrm{SB}=\mathrm{SC}$
Solving the equation $\mathrm{SA}=\mathrm{SB}=\mathrm{SC}$  . We get  $\left(\mathrm{\alpha },\mathrm{\beta }\right)=\left(\frac{-6}{7},\frac{2}{7}\right)$
Therefore,  ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=\frac{36}{49}+\frac{4}{49}=\frac{40}{49}=\boxed{0.816}$