2019/010) Given A, B, C, integers satisfying $A \log\, 16 + B \log\, 18 + C \log\, 24 = 0$ find minimum value of $A^2 + B^2+C^2$,

We are given
$A \log\, 16 + B \log\, 18 + C \log\, 24 = 0$
or
$A \log\, 2^4 + B \log\, 2* 3^2 + C \log\, 2^3*3 = 0$
or
$A (4\log\, 2) + B (\log\, 2 + 2\log\,  3) + C (3 \log\, 2 + \log\, 3) = 0$
or $\log\, 2( 4 A+ B + 3C) + \log\, 3(2B + C = 0$
as $\log\,2$ and $\log\,3$ are irrational so we must have

$4A + B + 3C = 0\cdots(1)$
$2B + C = 0\cdots(2)$
From (2)
$C = - 2B\cdots(3)$
And putting in (1) we get $4A= 5B\cdots(4)$

Lowest A, B , - C(all positive)  shall give lowest $A^2+B^2+C^2$
From (4) we get
$A=5, B= 4$ hence $C = -8$ from (3)

So lowest $A^2+B^2+C^2= 25 + 16 + 64 = 105$
(A= -5, B = -4, C= 8) gives same value