$a^3 \bmod 3^7$ depends only on $a \bmod 3^6,$ and cubes of numbers prime to $3$ are $\equiv \pm 1 \pmod 9$

Three units, or one unit, can’t sum to $0$ mod $9.$ With one multiple of $3,$ each valid target has exactly one cube root; with all three, recurse modulo $3^4.$

Since $(a + 3^6 t)^3 \equiv a^3 \pmod{3^7},$ the cube of $a$ modulo $3^7$ depends only on $a \bmod 3^6,$ and each residue occurs exactly once in $1 \le a \le 3^6.$ Moreover, the only cube roots of $1$ modulo $3^7$ are $1 + 3^6 t,$ which all agree modulo $3^6;$ hence cubing is a bijection from the $486$ units modulo $3^6$ onto the set of unit cubes modulo $3^7,$ which is exactly the set of units $\equiv \pm 1 \pmod 9.$ If all three of $a,$ $b,$ and $c$ are prime to $3$ (or exactly one is), then modulo $9$ the sum of cubes is $\pm 1 \pm 1 \pm 1$ or $\pm 1,$ never $0:$ no solutions.

Exactly one multiple of $3,$ say $c = 3z:$ for each of the $486$ units $a$ and $243$ choices of $c,$ the requirement $b^3 \equiv -a^3 - 27z^3 \pmod{3^7}$ has a right side that is a unit $\equiv \pm 1 \pmod 9,$ hence has exactly one solution $b$ modulo $3^6.$ With $3$ choices for which variable is the multiple of $3,$ this case gives $3 \cdot 486 \cdot 243 = 354294$ triples.

All three multiples of $3:$ writing $a = 3x$ etc. with $x,$ $y,$ and $z$ ranging modulo $3^5,$ the condition becomes $x^3 + y^3 + z^3 \equiv 0 \pmod{3^4},$ which depends only on the residues modulo $3^3,$ so the count is $9^3$ times the count modulo $27.$ Repeating the same analysis one level down: the two-unit case gives $3 \cdot 18 \cdot 9 = 486,$ and the all-divisible case reduces to $u + v + w \equiv 0 \pmod 3$ with $u,$ $v,$ and $w$ modulo $9,$ giving $243;$ that is $486 + 243 = 729$ triples modulo $27,$ hence $729 \cdot 729 = 531441$ here. In total $N = 354294 + 531441 = 885735,$ whose remainder modulo $1000$ is $735.$