Answer

Correct Option: (iii) Context-sensitive but not context-free (and hence not regular)

Explanation

$L = \{ a^n b^n c^{n+m} \mid m,n \ge 1 \} = \{ a^n b^n c^{k} \mid k = n+m > n \}$.

Not Context-Free

Assume $L$ is CFL, let $p$ be pumping length. Take $w = a^p b^p c^{p+1} \in L$ ($n=p, m=1$). By pumping lemma for CFL, $w = uvxyz$ with $|vxy|\le p$, $|vy|>0$. Pumping must stay within first $p$ $a$'s or $b$'s or $c$'s. Any pumping changes counts and breaks condition $\#_c > \#_a = \#_b$ (or breaks $a^n b^n$). Formal proof shows $L$ not CFL. Equivalent to $a^n b^n c^n$ (which is not CFL) with extra $c^m$ tail; intersection with regular $a^*b^*c^*$ does not help but $L \cap a^*b^*c^p$ style shows non-CFL.

Context-Sensitive

It is accepted by Linear Bounded Automaton. Grammar is context-sensitive:

$S \rightarrow a S B C \mid a B C$
$CB \rightarrow BC$
$a B \rightarrow a b$
$b B \rightarrow b b$
$b C \rightarrow b c$
$c C \rightarrow c c$

This is Type-1 (CSG) but not Type-2. It is decidable but not CFL.

Final Answer

CSL but not CFL → Option (iii). Also not regular; ultimately Type-1 in Chomsky hierarchy.