Step 1 — Analyze Statement I: \(a - b < c - a\)
Rewrite inequality:
\(a - b < c - a\)
\(\Rightarrow a + a < c + b\)
\(\Rightarrow 2a < b + c\).
Since \(a < 0\), the left-hand side \(2a\) is negative. The right-hand side \(b + c\) may be negative or positive depending on values.
Example: \(a = -5, b = -3, c = 2\). Then LHS = -2, RHS = 7 → inequality true.
But if \(a = -1, b = -100, c = 1\). Then LHS = -2, RHS = -99 → inequality false.
Therefore, Statement I is not always true.
Step 2 — Analyze Statement II: \(a + b < b + c\)
Simplify:
\(a + b < b + c\)
\(\Rightarrow a < c\).
Since \(a < 0\) and \(c > 0\), this is always true.
Hence, Statement II is always true.
Step 3 — Analyze Statement III: \(b < c < a\)
This requires \(c < a\). But \(c\) is positive and \(a\) is negative, so a positive number cannot be less than a negative number.
Therefore, Statement III is false.
Final Answer: Only Statement II is correct.