12. Let $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ be geometric sequences, and $c_{n}=a_{n}+b_{n}\left(n \geqslant 1, n \in \mathbf{N}_{+}\right)$.
(1) If $a_{n}=2^{n}, b_{n}=3^{n}$, and the sequence $\left\{c_{n+1}-p c_{n}\right\}$ is a geometric sequence, find the value of $p$;
(2) If the common ratios of the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are not equal, prove that the sequence $\left\{c_{n}\right\}$ is not a geometric sequence.
12. (1) Since $c_{2}-p c_{1}=13-5 p, c_{3}-p c_{2}=35-13 p, c_{4}-p c_{3}=97-35 p$, and $\left\{c_{n+1}-p c_{n}\right\}$ is a geometric sequence, we have $(13-5 p)(97-35 p)=(35-13 p)^{2}$, which simplifies to $p^{2}-5 p+6=0$. Solving this, we get $p=2, p=3$. Upon verification, both $p=2$ and $p=3$ make $\left\{c_{n+1}-p c_{n}\right\}$ a geometric sequence.
(2) Let the general term of $\left\{a_{n}\right\}$ be $a_{n}=a q_{1}^{n-1}$, and the general term of $\left\{b_{n}\right\}$ be $b_{n}=b q_{2}^{n-1}$, where $a \cdot b \neq 0, q_{1} \neq q_{2}$. Then, $c_{1} c_{3}-c_{2}^{2}=(a+b)\left(a q_{1}^{2}+b q_{2}^{2}\right)-\left(a q_{1}+b q_{2}\right)^{2}=a b\left(q_{1}-q_{2}\right)^{2} \neq 0$. Therefore, $c_{1} c_{3} \neq c_{2}^{2}$. Hence, the sequence $\left\{c_{n}\right\}$ is not a geometric sequence.