Given,
Binomial expression,
(2log2(10−3x)+52(x−2)log23)m
((10−3x)+53(x−2))m
Now, T6=C5m(10−3x)2m−5⋅(3x−2)=21…(1)
Also given,
C1m,C2m,C3m are in A.P.
So, 2⋅C2m=C1m+C3m
⇒2×2!(m−2)!m!=m+3!(m−3)!m!
Solving for m, we get m=2,7 and m=2 (rejected), so m=7
Put in equation (1)
21⋅(10−3x)93x=21
⇒(10−3x)3x=9×1
⇒3x=30,32
⇒x=0,2
Sum of the squares of all possible values of x=4.