option 2
This can be solved by sum and product of roots of cubic equation.
$Ax^3 +Bx^2 + Cx +D=0$ where pqr are three roots
so p*q*r = -D/A
p+q+r = -B/A
pq+qr+pr = C/A
for the given eqn $x^{3} - ax^{2} + bx - c=0$ consecutive roots are (k-1), k and (k+1)
So, (k-1)* k *(k+1) = +c
(k-1) + k + (k+1)=+a
k(k-1) + k(k+1) + (k-1)(k+1)=+b
=> 3k^2 -1 = b (here k will be alwys +ve)
b's minimum value is possible only when k is 0
and b= 3(0) - 1= - 1