Linearly independent vectors involved

Let $\displaystyle \vec{A} = \vec{a}+2\vec{b}-\vec{c} , \vec{B} = 2\vec{a}+\vec{b}+\vec{c} \text{ and } \vec{C} = 4\vec{a}-\vec{b}+\vec{c}$ . We want to find out the value of $\displaystyle \frac{[ \vec{A} \quad \vec{B} \quad \vec{C} ]}{[ \vec{a} \quad \vec{b} \quad \vec{c} ]}$ .

It can be easily noted that $2\vec{A} + \vec{C} = 3\vec{B}$ . And the scalar triple product of any three linearly dependent vectors is always $0$ . Also since $\displaystyle \vec{a} , \vec{b}$ and $\displaystyle \vec{c}$ are linearly independent their scalar triple product would be some non zero quantity. Hence, the answer is $\displaystyle \boxed{0}$ .

In JEE style : Since Answer is True for every Pair of a,b,c (put vectors on them)

So Let that $\displaystyle{ \vec{a} =\hat { i } \quad ,\quad \vec{ b } =\hat { j } \quad ,\quad \vec{ c } =\hat { k } \quad \\ { D }_{ r }=\left[ \begin{matrix} \vec { a } & \vec { b } & \vec { c } \end{matrix} \right] =1\\ \\ { N }_{ r }=\left| \begin{matrix} 1 & 2 & -1 \\ 2 & 1 & 1 \\ 4 & -1 & 5 \end{matrix} \right| =0\\ \boxed { Ans=0 } }$