= Solution
The <characteristic equation> has roots $-2,-6$, and a trial multiple of $e^{-4t}$ gives the <particular solution> $-3e^{-4t}$. Thus
$$
f(t)=Ae^{-2t}+Be^{-6t}-3e^{-4t}.
$$
The first two conditions give $A+B=3$ and $-2A-6B+12=0$, hence $A=B=3/2$. At $t_*=\log\sqrt2$, the exponentials $e^{-2t_*},e^{-4t_*},e^{-6t_*}$ equal $1/2,1/4,1/8$. Therefore
$$
f(t_*)=\frac34+\frac3{16}-\frac34=\frac3{16}\ne0.
$$
\b[No solution satisfies all three conditions.] The first two initial data already determine a unique solution of the <second-order linear differential equation>; the extra condition is inconsistent with it.
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