Find the circumcentre of the triangle whose sides are 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1= 0.

Given line are
3x – y – 5 = 0 ... (1)
x + 2y – 4 = 0 ... (2)
5x + 3y + 1 = 0 ... (3)
The point of intersection of (1) & (2) is A(2, 1)
The point of intersection of (2) & (3) is B(–2, 3)
The point of intersection of (3) &(1) is C(1, –2)
$\text{ Let }\mathrm{S}(\mathrm{\alpha },\mathrm{\beta })\text{ be circumcentre of the triangle }\mathrm{ABC}$
We know that SA = SB = SC
$\Rightarrow {\mathrm{SA}}^2={\mathrm{SB}}^2={\mathrm{SC}}^2$
$\text{ Now }{\mathrm{SA}}^2={\mathrm{SB}}^2$
$\begin{array}{l}\Rightarrow (\mathrm{\alpha }-2{)}^2+(\mathrm{\beta }-1{)}^2=(\mathrm{\alpha }+2{)}^2+(\mathrm{\beta }-3{)}^2\\ \Rightarrow {\mathrm{\alpha }}^2+{\mathrm{A}}^2-4\mathrm{\alpha }+{\mathrm{\beta }}^2+1-2\mathrm{\beta }\\ ={\mathrm{\alpha }}^2+\mathrm{\alpha ̸}+4\mathrm{\alpha }+{\mathrm{\beta }}^2+9-6\mathrm{\beta }\\ \Rightarrow 8\mathrm{\alpha }-4\mathrm{\beta }+8=0\\ 2\mathrm{\alpha }-\mathrm{\beta }+2=0 ..... \left(4\right)\end{array}$
$\text{ And }{\mathrm{SB}}^2={\mathrm{SC}}^2$
$\begin{array}{l}\Rightarrow (\mathrm{\alpha }+2{)}^2+(\mathrm{\beta }-3{)}^2=(\mathrm{\alpha }-1{)}^2+(\mathrm{\beta }+2{)}^2\\ \Rightarrow {\mathrm{\alpha }}^2+4+4\mathrm{\alpha }+{\mathrm{\beta }}^2+9-6\mathrm{\beta }\\ ={\mathrm{\alpha }}^2+1-2\mathrm{\alpha }+{\mathrm{\beta }}^2+4+4\mathrm{\beta }\\ \Rightarrow 6\mathrm{\alpha }-10\mathrm{\beta }+8=0\\ \Rightarrow 3\mathrm{\alpha }-5\mathrm{\beta }+4=0\dots .\left(5\right)\end{array}$
Solve (4) & (5)
$\begin{array}{l} \mathrm{\alpha }\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\beta }\hspace{0.25em} 1\\ -1\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-1\\ -5\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} -5\end{array}$
$\frac{\mathrm{\alpha }}{6}=\frac{\mathrm{\beta }}{-2}=\frac{1}{-7};\mathrm{\alpha }=\frac{-6}{7},\mathrm{\beta }=\frac{2}{7}$
$\therefore \text{ Circumcentre }\mathrm{S}(\mathrm{\alpha },\mathrm{\beta })=\left(\frac{-6}{7},\frac{2}{7}\right)$