The given equation of lines are
y = m1x+c1 ... (i)
y = m2x+c2 ... (ii)
y = m3x+c3 ... (iii)
On solving (i) and (ii), we get x =

−(c2−c1)

m2−m1

, y =

c1m2−c2m1

m2−m1

Hence, (i) & (ii) meets at [

−(c1−c2)

(m1−m2)

,

(m1c2−m2c1)

(m1−m2)

]
Clearly (i), (ii) and (iii) lines are concurrent if the above point lies on (iii)
i.e., if [

m1c2−m2c1

m1−m2

] = m2[

−(c1−c2)

(m1−m2)

]+c3
i.e., if

(m1c2−m2c1)

(m1−m2)

=

m3(−(c1−c2))

(m1−m2)

+c3
i.e., if (m1c2−m2c1) = m3(c2−c1)+c3(m1−m2)
⇒ m1c2–m2c1 = m3c2–m3c1+c3m1–c3m2
⇒ m1c2–m2c1–m3c2+m3c1–c3m1+c3m2 = 0
⇒ m1(c2–c3)+m2(c3–c1)+m3(c1–c2) = 0.
Hence proved.