The equation of the given lines are :

$ \qquad y = m_1x+c_1$------(1)

$ \qquad y = m_2x+c_2$--------(2)

$ \qquad y=m_3x+c_3$-------(3)

Subtract equation (1) and (2) we get,

$ 0= (m_2-m_1)x+(c_2-c_1)$

$ \Rightarrow x = \large\frac{c_2-c_1}{m_1-m_2}$

Substituting the value of $x$ in (1) we get,

$ y = m_1 \bigg( \large\frac{c_2-c_1}{m_1-m_2} \bigg) $$+c_1$

$ \Rightarrow y = \large\frac{m_1c_2-m_1c_1}{m_1-m_2}$$+c_1$

$ y =\large\frac{m_1c_2-m_1c_1+m_1c_1-m_2c_2}{m_1-m_2}$

$ \Rightarrow y = \large\frac{m_1c_2-m_2c_1}{m_1-m_2}$

Hence the point of intersection is $ \bigg( \large\frac{c_2-c_1}{m_1-m_2}$$, \large\frac{m_1c_2-m_2c_1}{m_1-m_2} \bigg)$

Since the line (1), (2) and (3) are concurrent, the above point of intersection should also satisfy the equation (3).

Hence substituting for $x$ and $y$ we get,

$ \large\frac{m_1c_2-m_2c_1}{m_1-m_2}$$ = m_3 \bigg( \large\frac{c_2-c_1}{m_1-m_2} \bigg)$$+c_3$

On simplifying we get,

$\large\frac{m_1c_2-m_2c_1}{m_1-m_2}$$ = \large\frac{m_3c_2-m_3c_1+c_3m_1-c_3m_2}{m_1-m_2}$

$ \Rightarrow m_1c_2-m_2c_1-m_3c_2-c_3m_1+m_2c_3=0$

$ \Rightarrow m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0$

Hence proved.