Combining Unit Fractions!

Relevant wiki: General Diophantine Equations - Problem Solving

The equation can be rearranged to $42(ab+ac+bc) = abc$ .

WLOG, assuming $a$ divisible by 7, we set $a = 7A$ are: $Abc = 6 * (7A * (b + c) + bc )$ .

$bc * (A-6) = 6 * 7A * (b + c)$

Go into one case assuming $A-6$ divisible by 7. Set the $A - 6 = 7t$

$\Rightarrow$ $A = 7 * (7t + 6) = 49t + 42$

$bct = 6 * (7t + 6) * (b + c)$

Choose $t = 1$ , then $a = 49 + 42 = 91$

$bc = 6 * 13 * (b+c)$ , to assume $b$ is divisible by 13, set $b = 13B$ $\Rightarrow$ $Bc = 6 * (13B + c )$

$\Rightarrow$ $(B - 6) * c = 6 * 13 * B$

Again assuming $B - 6$ is divisible by 13, then $B - 6 = 13m$ $\Rightarrow$ $mc = 6 * (13m + 6)$ ,

$m = 1, c = 114, b = 247$

If we choose $t = 2$ , $\Rightarrow$ $a = 49 * 2 + 42 = 140$ , $2bc = 6 * 20 * (b + c)$

$bc = 60 * (b + c) = 3 * 4 * 5 * (b + c)$

Suppose $b$ is divisible by 5, set $b = 5B$ $\Rightarrow$ $Bc = 3 * 4 * (5B + c)$

$\Rightarrow$ $C * (B - 12) = 3 * 4 * 5B$ $\Rightarrow$ $c = 60+ 720 / (B - 12)$

Get the $B - 12 = 10$ $\Rightarrow$ $c = 132$ $\Rightarrow$ $b = 110$

So, $\min(a+b+c) = 382.$

1/a + 1/b + 1/c = 1/42,
(a + b + c) shall be minimum when a=b=c=3×42=126,
If one of the number, say a, is kept at 126, then, we have,
1/b + 1/c = (1/42) - (1/126) = 1/63,
63(b + c) = bc, or b(c - 63) = 63c,
b = 63 + (63×63)/(c - 63),
If (c - 63) = 63, then b=c=a=126, which is not permissible because a,b and c are different. Therefore, (c - 63) could be 7×7 or 9×9 so as to keep c and b close to 126.
c - 63 = 49 or 81, gives c = 112 or 144, and
b = 144 or 112.
So, a + b + c = 126 + 112 + 144 = 382.
Other triplets giving the same minimum value of 382 are also possible if a is not assumed to be 126, but is taken as having a factor in common with 42.