Question 696813
2a+3b+c=11  ---------------------------------eq.1
6ab+2ac+3bc=24-------------------------------eq.2
abc=-6  -------------------------------------eq.3

From Eq.1 ,  2a+3b = 11 - c
From eq.2 , 6ab + c (11 - c) = 24
{{{6ab+11c-c^2 = 24}}}----------------------eq.4

From eq.3 , ab = -6/c


Therefore eq.4 becomes
{{{6(-6/c)+11c-c^2 = 24}}}
{{{-36+11c^2-c^3 = 24c}}}

{{{c^3-11c^2+24c+36 = 0}}} is a cubic equation 
By trial and error , one of the roots of the equation is -1.
Therefore ,(c-1) is a factor.
Let us reduce the cubic equation to a quadratic equation.
We get , {{{c^2-12c+36}}} is the quotient which is a quadratic equation.

The roots are 6 and 6.
c = {-1,6}

Therefore 2a+3b = 11-(-1) = 12 --------------------eq.5
ab(-1) = -6
ab = 6
b  = 6/a

Substitute for b in eq.5

{{{2a+3(6/a) = 12}}}
{{{2a^2+18 -12a = 0}}} is a quadratic equation whose roots are 3 and 3

Therefore , b = 2

(a,b,c) = (3,2,-1)  when c = -1
But when  c = 6, we have 2 possibilities for a and b

(a,b,c) = (3,-1/3,6)  OR (-1/2 , 2,6)