if x^2-2ra[r]x+r=0,r=1,2,3 are three quadratic equations of which each pair has exactly one root common then the number of solutions of the triplet (a1,a2,a3) is
Your notation is a bit strange. Do you mean $\displaystyle x^2- 2a_1x+ 1= 0$, $\displaystyle x^2- 4a_2x+ 2= 0$, and $\displaystyle x^2- 6a_3x+ 3= 0$?
By the quadratic formula, the solution to the first is $\displaystyle x= a_1\pm\sqrt{a_1^2- 1}$, to the second $\displaystyle x= 2a_2\pm\sqrt{4a_2^2- 2}$, and to the third $\displaystyle x= 3a_3\pm\sqrt{9a_3- 3}$. Compare those in pairs so see how you can get one common root.