Can you solve 5a greatly appreciated.

Part I)
 The module of vector AB is given by:
 lABl = root ((- 3) ^ 2 + (4) ^ 2)
 lABl = root (9 + 16)
 lABl = root (25)
 lABl = 5

 Part (ii)
 The module of the EF vector is given by:
 lEFl = root ((5) ^ 2 + (e) ^ 2)
 We have to:
 lEFl = 3lABl
 Thus:
 root ((5) ^ 2 + (e) ^ 2) = 3 * (5)
 root ((5) ^ 2 + (e) ^ 2) = 15
 Clearing e have:
 (5) ^ 2 + (e) ^ 2 = 15 ^ 2
 (e) ^ 2 = 15 ^ 2 - 5 ^ 2
 e = root (200)
 e = root (2 * 100)
 e = 10 * root (2)

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.