Structural determination of carbohydrates(sucrose)
First of all, lets be familiar with what per-iodic acid does to organic compounds.

It selects the carbon-carbon bonds that have either an -OH group, a =O(carbonyl) group or if it is the functional group carbon of a carboxylic acid. Both the carbons of the C-C bond must satisfy this condition. Here are some examples of the selection points of per-iodic acid:


Now that per-iodic acid has chosen the center, it cleaves the C-C bond and adds an -OH at each carbon. For every C-C bond, one per-iodic acid molecule is consumed. I am avoiding going throughout the actual mechanism to keep things simple. If any of the carbons has two -OH groups, they will spontaneously dehydrated to give a =O bond instead. For the examples shown above, the final products of the cleavage-oxidation would be:


Now, coming to your main question, the structure of sucrose would be:


By what we have seen, lets mark the C-C bonds that per-iodic acid can cleave:


Since we have 3 centers, so thats why 3 molecules of per-iodic acid are consumed per molecule of sucrose. Now you can even proceed and find out the oxidation products.

I hope this answer helps. Happy carbohydrate breaking!