smc

Assume $c,d\ge 0$, and WLOG, assume $a<0$ and $a\le b$. This also takes into account when $b$ is negative. That means $$\frac{1}{2^{-a}}+2^b=3^c+3^d$$ Multiply both sides by $2^{-a}$ to get $$1+2^{-a+b}=2^{-a}(3^c+3^d)$$ Note that both sides are integers. If $a\ne b$, then the right side is even while the left side is odd, so equality can not happen. If $a=b$, then $2^{-a}(3^c+3^d)=2$, and since $a<0$, $a=-1$ and $3^c+3^d=1$. No nonnegative value of $c$ and $d$ works, so equality can not happen. Thus, $a$ and $b$ can not be negative when $c,d\ge 0$. Assume $a,b\ge 0$, and WLOG, assume $c<0$ and $c\le d$. This also takes into account when $d$ is negative. That means $$2^a+2^b=\frac{1}{3^{-c}}+3^d$$ Multiply both sides by $3^{-c}$ to get $$3^{-c}(2^a+2^b)=1+3^{d-c}$$ That makes both sides integers. The left side is congruent to $0$ modulo $3$ while the right side is congruent to $1$ or $2$ modulo $3$, so equality can not happen. Thus, $c$ and $d$ can not be negative when $a,b\ge 0$. Assume $a,c<0$, and WLOG, let $a\le b$ and $c\le d$. This also takes into account when $b$ or $d$ is negative. That means $$\frac{1}{2^{-a}}+2^b=\frac{1}{3^{-c}}+3^d$$ Multiply both sides by $2^{-a}\cdot 3^{-c}$ to get $$3^{-c}(1+2^{b-a})=2^{-a}(1+3^{d-c})$$ That makes both sides integers. The left side is congruent to $0$ modulo $3$ while the right side is congruent to $1$ or $2$ modulo $3$, so equality can not happen. Thus, $a$ and $c$ can not be negative. Putting all the information together, none of $a,b,c,d$ can be negative, so the answer is $\boxed{0}$.