For each bit, ($a^$b^$c) returns

• 1 if all three cards share the value represented by that bit (e.g. all three cards are green),
• 0 if exactly two of three cards share the value represented by that bit (e.g. two of the three cards are green, and the one others isn't),
• 1 if exactly one of three cards share the value represented by that bit (e.g. one of the three cards is green, and the two others aren't),
• 0 if none of the three cards share the value represented by that bit (e.g. none of the cards are green).

We want

• 1 if all three cards share the value represented by that bit (e.g. all three cards are green),
• 0 if exactly two of three cards share the value represented by that bit (e.g. two of the three cards are green, and the one others isn't),
• 1 if exactly one of three cards share the value represented by that bit (e.g. one of the three cards is green, and the two others aren't),
• 1 if none of the three cards share the value represented by that bit (e.g. none of the cards are green).

($a|$b|$c) eq serves that purpose and to "and" the results of every bit into a single boolean value.