Correct Answer - D
We have,
`("log"_(2)a)/(2) = ("log"_(3)b)/(3) = ("log"_(4)c)/(4) = lambda ("say")`
`rArr a = 2^(2lambda), b = 3^(3 lambda), c = 4^(4lambda)`
Now,
`a^(1//2) b^(1//3) c^(1//4) = 24`
`rArr 2^(lambda) xx 3^(lambda) xx 4^(lambda) = 24 rArr 2^(lambda) xx 3^(lambda) xx 4^(lambda) = 2 xx 3 xx 4 rArr lambda = 1`
Hence, `a=2^(2) = 4, b=3^(3) = 27 " and " c = 4^(4) = 256`
