Let ABC be a triangle of area $$15\sqrt{2}$$ and the vectors $$\overrightarrow{AB} = \hat{i} + 2\hat{j} - 7\hat{k}$$, $$\overrightarrow{BC} = a\hat{i} + b\hat{j} + c\hat{k}$$ and $$\overrightarrow{AC} = 6\hat{i} + d\hat{j} - 2\hat{k}$$, d > 0. Then the square of the length of the largest side of the triangle ABC is ______.

AB = i + 2j - 7k, AC = 6i + dj - 2k. BC = AC - AB = 5i + (d-2)j + 5k.

Area = (1/2)|AB × AC| = 15√2, so |AB × AC| = 30√2.

AB × AC = |i j k; 1 2 -7; 6 d -2| = (2(-2)-(-7)d)i - (1(-2)-(-7)(6))j + (1d-2(6))k

= (-4+7d)i - (-2+42)j + (d-12)k = (7d-4)i - 40j + (d-12)k

|AB × AC|² = (7d-4)² + 1600 + (d-12)² = 49d²-56d+16+1600+d²-24d+144 = 50d²-80d+1760

= (30√2)² = 1800

50d²-80d+1760 = 1800 → 50d²-80d-40 = 0 → 5d²-8d-4 = 0

(5d+2)(d-2) = 0 → d = 2 (since d > 0)

|AB|² = 1+4+49 = 54, |BC|² = 25+0+25 = 50, |AC|² = 36+4+4 = 44

Largest side: AB² = 54

The answer is 54.