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If A = 1/5! 6! 7! [(5!, 6!, 7!),(6!, 7!, 8!),(7!, 8!, 9!)], then

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If A = \(\frac{1}{5!6!7!}\begin{bmatrix} 5! & 6! &7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{bmatrix}\) then adj(adj(2A)) is equal to

(1) 28

(2) 212 

(3) 220

(4) 216

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Correct option is (4) 216

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