Let $$\vec{b} = \hat{i} + \hat{j} + \lambda\hat{k}, \lambda \in \mathbb{R}$$. If $$\vec{a}$$ is a vector such that $$\vec{a} \times \vec{b} = 13\hat{i} - \hat{j} - 4\hat{k}$$ and $$\vec{a} \cdot \vec{b} + 21 = 0$$, then
$$\vec{b} - \vec{a} \cdot \hat{k} - \hat{j} + \vec{b} + \vec{a} \cdot \hat{i} - \hat{k}$$ is equal to ______

We have $$\vec{b} = \hat{i} + \hat{j} + \lambda\hat{k}$$, $$\vec{a} \times \vec{b} = 13\hat{i} - \hat{j} - 4\hat{k}$$, and $$\vec{a} \cdot \vec{b} + 21 = 0$$. Let $$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$. Computing $$\vec{a} \times \vec{b}$$ gives $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 1 & 1 & \lambda \end{vmatrix} = (a_2\lambda - a_3)\hat{i} - (a_1\lambda - a_3)\hat{j} + (a_1 - a_2)\hat{k},$$ and equating this with $$13\hat{i} - \hat{j} - 4\hat{k}$$ yields $$a_2\lambda - a_3 = 13 \quad\cdots(1),$$ $$-(a_1\lambda - a_3) = -1 \implies a_3 - a_1\lambda = -1 \quad\cdots(2),$$ $$a_1 - a_2 = -4 \quad\cdots(3).$$

Adding (1) and (2) gives $$(a_2\lambda - a_3)+(a_3 - a_1\lambda)=13+(-1)\implies \lambda(a_2 - a_1)=12.$$ Since (3) gives $$a_2 - a_1=4$$, it follows that $$4\lambda=12$$ and hence $$\lambda=3$$.

The dot product condition $$\vec{a}\cdot\vec{b}=a_1+a_2+3a_3=-21\quad\cdots(4)$$ provides another relation. From (3) we have $$a_1=a_2-4$$, and substituting $$\lambda=3$$ into (1) gives $$3a_2 - a_3=13\implies a_3=3a_2-13$$. Substituting these into (4) yields $$(a_2 - 4)+a_2+3(3a_2 -13)=-21,$$ which simplifies to $$a_2 - 4 + a_2 +9a_2 -39=-21\implies 11a_2 -43=-21\implies 11a_2=22\implies a_2=2.$$ Therefore $$a_1=2-4=-2$$ and $$a_3=3\cdot2-13=-7$$, giving $$\vec{a}=-2\hat{i}+2\hat{j}-7\hat{k}$$ and $$\vec{b}=\hat{i}+\hat{j}+3\hat{k}$$.

It can be verified that $$\vec{a}\times\vec{b}=(2\cdot3-(-7))\hat{i}-((-2)\cdot3-(-7))\hat{j}+((-2)-2)\hat{k}=13\hat{i}-\hat{j}-4\hat{k},$$ and $$\vec{a}\cdot\vec{b}=-2+2-21=-21,$$ so indeed $$\vec{a}\cdot\vec{b}+21=0$$.

Finally, to compute $$(\vec{b}-\vec{a})\cdot(\hat{k}-\hat{j})+(\vec{b}+\vec{a})\cdot(\hat{i}-\hat{k}),$$ note that $$\vec{b}-\vec{a}=(1-(-2))\hat{i}+(1-2)\hat{j}+(3-(-7))\hat{k}=3\hat{i}-\hat{j}+10\hat{k},$$ so $$(\vec{b}-\vec{a})\cdot(\hat{k}-\hat{j})=0+1+10=11.$$ Also, $$\vec{b}+\vec{a}=(1+(-2))\hat{i}+(1+2)\hat{j}+(3+(-7))\hat{k}=-\hat{i}+3\hat{j}-4\hat{k},$$ and $$(\vec{b}+\vec{a})\cdot(\hat{i}-\hat{k})=-1+0+4=3.$$ Therefore the total is $$11+3=14\text{.}$$

The answer is $$\boxed{14}$$.