Discussion

Explanation:

Let AB = a3 and AC = 3n

Also

AC = 3 × AB

∴ 3n = 3 × a3

∴ 3(– 1) = a3

Now, let the perimeter be equal to p

= BC + AC + AB

= 397 + 3n + 3(– 1)

(p – 397) = 3(– 1) × (3 + 1)

= 3(– 1) × 4

Thus the LHS of the above equation should be a multiple of 3 and 4. Substitute the value of perimeter given in the options and verify this. Among the options, only (3313 – 397) is divisible by 3 and 4..

Hence, option (d).

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