Let $$\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}$$, $$\vec{b} = 4\hat{i} + \hat{j} + 7\hat{k}$$ and $$\vec{c} = \hat{i} - 3\hat{j} + 4\hat{k}$$ be three vectors. If a vector $$\vec{p}$$ satisfies $$\vec{p} \times \vec{b} = \vec{c} \times \vec{b}$$ and $$\vec{p} \cdot \vec{a} = 0$$, then $$\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k})$$ is equal to

$$\vec{p}\times\vec{b}=\vec{c}\times\vec{b}$$, so $$(\vec{p}-\vec{c})\times\vec{b}=0$$, meaning $$\vec{p}-\vec{c}=\lambda\vec{b}$$.

$$\vec{p}=\vec{c}+\lambda\vec{b}=(1+4\lambda)\hat{i}+(-3+\lambda)\hat{j}+(4+7\lambda)\hat{k}$$.

$$\vec{p}\cdot\vec{a}=0$$: $$3(1+4\lambda)+1(-3+\lambda)-2(4+7\lambda)=0$$.

$$3+12\lambda-3+\lambda-8-14\lambda=0 \Rightarrow -\lambda-8=0 \Rightarrow \lambda=-8$$.

$$\vec{p}=(-31)\hat{i}+(-11)\hat{j}+(-52)\hat{k}$$.

$$\vec{p}\cdot(\hat{i}-\hat{j}-\hat{k})=-31+11+52=32$$.

The answer is Option (4): $$\boxed{32}$$.