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**Hint:**Solve the given equations pairwise to obtain the vertices of the triangle. Substitute the coordinates of the vertices in the general equation of a circle ${(x - h)^2} + {(y - k)^2} = {r^2}$ with $(h,k)$ as its centre. Solve the equations thus obtained to get $(h,k)$ which is the required answer.

**Complete step by step solution:**

We are given the equation of the lines which form the sides of the triangle.

They are $3x - y - 5 = 0$, $x + 2y - 4 = 0$, $5x + 3y + 1 = 0$.

We need to find the circumcenter of the triangle.

In the above figure, consider O as the circumcenter of $\vartriangle ABC$.

Consider the system of linear equations

$3x - y - 5 = 0......(1)$

$x + 2y - 4 = 0.....(2)$

$5x + 3y + 1 = 0....(3)$

As can be understood from the figure, every pair of lines has a point of intersection and these 3 points of intersection form the vertices of the triangle.

From (1), we get \[3x - 5 = y \Rightarrow y = 3x - 5\]

From (2), we get $2y = - x + 4 \Rightarrow y = - \dfrac{x}{2} + 2$

Therefore,

\[

3x - 5 = - \dfrac{x}{2} + 2 \\

\Rightarrow \dfrac{{7x}}{2} = 7 \\

\Rightarrow x = 2 \\

\]

Substituting\[x = 2\]in\[y = 3x - 5\], we get\[y = 3(2) - 5 = 1\]

Thus, we get one of the vertices of the triangle. Call this vertex as A.

$A \equiv (2,1)$

Similarly, by solving the remaining equations in pairs, we get the other two vertices.

$B \equiv (1, - 2)$ and $C \equiv ( - 2,3)$

Now, these vertices of $\vartriangle ABC$lie on the circle and we need only 3 points to construct a circle.

The centre of the circle formed by the vertices A, B, C is the required circumcenter.

We will construct the equation of the circle using A, B, and C.

Let $(h,k)$be the circumcenter.

Then the equation of the circle with radius r and centre $(h,k)$ is given by ${(x - h)^2} + {(y - k)^2} = {r^2}....(4)$

As the points A, B, and C lie on the circle, they will satisfy equation (4).

$A \equiv (2,1) \Rightarrow {(2 - h)^2} + {(1 - k)^2} = {r^2}.....(5)$

$B \equiv (1, - 2) \Rightarrow {(1 - h)^2} + {( - 2 - k)^2} = {r^2}.....(6)$

$C \equiv ( - 2,3) \Rightarrow {( - 2 - h)^2} + {(3 - k)^2} = {r^2}....(7)$

Comparing the equations (5) and (6), we get

\[

{(2 - h)^2} + {(1 - k)^2} = {(1 - h)^2} + {( - 2 - k)^2} \\

\Rightarrow 4 - 4h + {h^2} + 1 - 2k + {k^2} = 1 - 2h + {h^2} + 4 + 4k + {k^2} \\

\Rightarrow - 2h - 6k = 0 \\

\Rightarrow 2h + 6k = 0....(8) \\

\]

Similarly, from (6) and (7), we get

$3h - 5k = - 4.....(9)$

Solving (8) and (9), we get $(h,k) \equiv (\dfrac{{ - 6}}{7},\dfrac{2}{7})$

Hence, the circumcenter is $(\dfrac{{ - 6}}{7},\dfrac{2}{7})$.

**Note:**In the given $\vartriangle ABC$, the line passing through the circumcenter and the vertex A is the perpendicular bisector of its opposite side BC. This holds true for all the vertices.

In fact, the circumcenter of a triangle is defined as the point of intersection of the perpendicular bisectors of the sides of the triangle.

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