Sum of 3 Unit Fractions that equals 1

Theorem
There are $3$ ways to represent $1$ as the sum of exactly $3$ unit fractions.

Proof
Let:
 * $1 = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c$

where:
 * $0 < a \le b \le c$

and:

$a = 1$.

Then:
 * $1 = \dfrac 1 1 + \dfrac 1 b + \dfrac 1 c$

and so:
 * $\dfrac 1 b + \dfrac 1 c = 0$

which contradicts the stipulation that $b, c > 0$.

So there is no solution possible when $a = 1$.

Therefore $a \ge 2$.

$a = 2$
Let $a = 2$.

$b = 2$
Let $b = 2$.

Then:
 * $\dfrac 1 a + \dfrac 1 b = 1$

leaving no room for $c$.

Hence there are no solutions where $a = 2$ and $b = 2$.

$b = 3$
Let $b = 3$.

Thus we have:
 * $(1): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 6$

$b = 4$
Let $b = 4$.

Thus we have:
 * $(2): \quad 1 = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 4$

$b > 4$
Let $b > 4$.

Then:

Hence there are no solutions such that $a = 2, b > 4$.

$a = 3$
Let $a = 3$.

$b = 3$
Let $b = 3$.

Thus we have:
 * $(3): \quad 1 = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 3$

$a > 3$
Let $a > 3$.

Then:

Hence there are no solutions for $a>3$.

Summary
Hence our $3$ solutions: