Write $\overline{abc}=100a+10b+c$ with $1\le a\le 9$, $0\le b\le 9$, $1\le c\le 9$ (the last because $\overline{cba}$ is a three-digit number).

Condition (iii): $\overline{cba}>\overline{abc} \Leftrightarrow 99(c-a)>0 \Leftrightarrow c>a.$

Condition (i) (divisibility by $11$): $a - b + c \equiv 0\pmod{11}.$ Since $a,b,c\in[0,9]$, $a-b+c\in[-9,18]$. The achievable multiples of $11$ are $0$ and $11.$

Case 1: $a-b+c=0$, i.e., $b=a+c.$

Combine with condition (ii): $a+b+c=14 \Rightarrow 2(a+c)=14 \Rightarrow a+c=7,\ b=7.$

With $c>a$, $a\ge 1$, $c\le 9$: pairs $(a,c)$ are $(1,6),(2,5),(3,4)$.

Numbers: $176,\ 275,\ 374.$

Case 2: $a-b+c=11$, i.e., $b = a+c - 11.$

Combine: $a+b+c = 2(a+c)-11 = 14 \Rightarrow a+c = 12.5$, not an integer. No solutions.

Sum $= 176+275+374 = 825.$

Why this is CAT-level: Three conditions interact non-trivially. Condition (iii) -- expressed as ``$\overline{cba}-\overline{abc}$ is a positive multiple of $99$'' -- looks restrictive but is automatic for any reversal; the actual content is just $c>a$. Recognising this collapse is the key insight. The divisibility rule for $11$ produces a case split (Case 2 is killed by parity).

Answer: $176+275+374 = 825.$