How would you solve this system if linear equations in 3 variables? The equations are-
$$
a+b+c=6 
$$
$$
ab+bc+ca=11
$$
$$
abc=6
$$
Also,is there a way to generalise this to a formula for varying values of the constants?
Hint: $a,b,c$ are roots of the cubic equation
$$
x^3-6x^2+11x-6=0.
$$
An elementary computation (to get to the same end result as in the same answers):

Use the first equation to find that 
$$
a=6-b-c.
$$

Substitute this into the second equation and solve for $b$ using the quadratic formula to get
$$
b=-\frac{1}{2}c+3\pm\frac{1}{2}\sqrt{-3c^2+12c-8}
$$

Substituting this into the third equation results in
$$
c^3-6c^2+11c=6
$$
(which then must be solved).