Prove that (vec a×vec b)×(vec c×vec d)=[(vec d×vec a)·vec b]…
Prove that $(\vec a\times\vec b)\times(\vec c\times\vec d)=[(\vec d\times\vec a)\cdot\vec b]\,\vec c-[(\vec a\times\vec b)\cdot\vec c]\,\vec d$.
Explanation of Method
The vector product of two vector products.
Step 1: Proof
With $\vec m=\vec a\times\vec b$ and "BAC–CAB": \[\vec m\times(\vec c\times\vec d)=\vec c\,(\vec m\cdot\vec d)-\vec d\,(\vec m\cdot\vec c)=[\vec a,\vec b,\vec d]\,\vec c-[\vec a,\vec b,\vec c]\,\vec d.\]
By a cyclic shift, $[\vec a,\vec b,\vec d]=[\vec d,\vec a,\vec b]=(\vec d\times\vec a)\cdot\vec b$: \[(\vec a\times\vec b)\times(\vec c\times\vec d)=[\vec a,\vec b,\vec d]\,\vec c-[\vec a,\vec b,\vec c]\,\vec d\qquad\blacksquare\]
Step 2: Second form
Grouping instead with $\vec n=\vec c\times\vec d$ and $(\vec a\times\vec b)\times\vec n=\vec b(\vec a\cdot\vec n)-\vec a(\vec b\cdot\vec n)$: \[(\vec a\times\vec b)\times(\vec c\times\vec d)=[\vec a,\vec c,\vec d]\,\vec b-[\vec b,\vec c,\vec d]\,\vec a.\]
The result lies both in the plane of $\vec c,\vec d$ and in the plane of $\vec a,\vec b$: it is along their line of intersection.
Numerical check of the main form: $(\vec d\times\vec a)\cdot\vec b=1$ and $(\vec a\times\vec b)\cdot\vec c=1$, so the right side is $\vec c-\vec d=\{-1,-1,0\}$, which equals $\{0,0,1\}\times\{-1,1,0\}$ $\checkmark$
Result
\[\boxed{\,(\vec a\times\vec b)\times(\vec c\times\vec d)=[\vec a,\vec b,\vec d]\,\vec c-[\vec a,\vec b,\vec c]\,\vec d\qquad\blacksquare\,}\]
Remark
Sign correction. The source's second form has the signs reversed: $[\vec b,\vec c,\vec d]\,\vec a-[\vec a,\vec c,\vec d]\,\vec b$ (reached after a "Wait"). The source's own example shows this: with $\vec a=\vec i$, $\vec b=\vec j$, $\vec c=\vec k$, $\vec d=\{1,1,1\}$ we have $[\vec b,\vec c,\vec d]=1$ and $[\vec a,\vec c,\vec d]=-1$, so the source's form gives $\{1,1,0\}$, while the true value is $\{-1,-1,0\}$. The form above gives $-\vec j-\vec i=\{-1,-1,0\}$ $\checkmark$