Let $a,b, c$ denote the real numbers such that $1 \le  a, b, c\le  2$. 
Consider $T = (a - b)^{2018} + (b - c)^{2018} + (c - a)^{2018}$.
Determine the largest possible value of $T$.
1. Given the conditions \(1 \le a, b, c \le 2\), we need to determine the largest possible value of \(T = (a - b)^{2018} + (b - c)^{2018} + (c - a)^{2018}\).

2. First, note that \(|a - b| \le 1\), \(|b - c| \le 1\), and \(|c - a| \le 1\) because \(a, b, c\) are all within the interval \([1, 2]\).

3. Since \(2018\) is an even number, \((x)^{2018} = |x|^{2018}\). Therefore, we can rewrite \(T\) as:
   \[
   T = |a - b|^{2018} + |b - c|^{2018} + |c - a|^{2018}
   \]

4. To maximize \(T\), we need to consider the maximum values of \(|a - b|\), \(|b - c|\), and \(|c - a|\). Since \(|a - b|, |b - c|, |c - a| \le 1\), the maximum value of \(|x|^{2018}\) for any \(x\) in \([-1, 1]\) is \(1\).

5. We need to check if it is possible for \(T\) to achieve the value of \(2\). Assume \(a, b, c\) are such that the differences \(|a - b|\), \(|b - c|\), and \(|c - a|\) are maximized. Without loss of generality, assume \(a \ge b \ge c\).

6. If \(a = 2\), \(b = 1\), and \(c = 1\), then:
   \[
   |a - b| = |2 - 1| = 1, \quad |b - c| = |1 - 1| = 0, \quad |c - a| = |1 - 2| = 1
   \]

7. Substituting these values into \(T\), we get:
   \[
   T = 1^{2018} + 0^{2018} + 1^{2018} = 1 + 0 + 1 = 2
   \]

8. Therefore, the maximum value of \(T\) is indeed \(2\).

The final answer is \(\boxed{2}\).