Q.$\vec a=2\hat i+3\hat j-\hat k,\ \vec b=-\hat i+2\hat j-4\hat k,\ \vec c=\hat i+\hat j+\hat k$, then find the value of $(\vec a\times\vec b)\cdot(\vec a\times\vec c)$.

Two master identities connect FOUR vectors at once, both proved from the vector-triple-product expansion $\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c$.

Jacobi's identity. For any three vectors $\vec a,\vec b,\vec c$:

Expanding all three terms and using $\vec x\cdot\vec y=\vec y\cdot\vec x$, every term cancels a partner exactly — the three "cyclic" vector triple products of any three vectors always balance. This is what lets you compute an unknown fourth vector triple product from the other three, or prove a combination is automatically $\vec 0$ without any component bookkeeping.

Lagrange's identity. For any FOUR vectors $\vec a,\vec b,\vec c,\vec d$:

It converts the dot product of two cross products — normally requiring you to compute both cross products first — into just FOUR ordinary dot products, arranged as a $2\times2$ determinant. Setting $\vec c=\vec a,\vec d=\vec b$ recovers the familiar $|\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2$. …