(a) $10X + 12Y − 8Z ≡ 14 \pmod 8$
(b) $3X − 15Y + 24Z ≡ 13 \pmod {81}$
So I see this is a system of equations and I take the equations and did:
$(b)-3(a)$ which yields:
$33X +21Y ≡ -29 \pmod {81-8}$ ?
I'm unsure of if I can directly compute $\mod{(81-8)}$ however. Ideally I would want both mods to be the same number I believe.
$(b)$ does not have a solution in integers, since $3X - 15Y + 24Z$ is always a multiple of $3$, so the remainder when it is divided by $81$(which is a multiple of three) must also be a multiple of $3$. However, that number is given to be $13$, which is not a multiple of three.
On the other hand, $(a)$ has a solution in integers when considered in isolation : just take $X = Y=Z = 1$, for example. Naturally, both $(a)$ and $(b)$ can't be satisfied simultaneously, as $b$ can never be satisfied.
On a general note, if you are given two congruences mod different numbers, then you cannot add these congruences.