If $$\begin{bmatrix} \vec{a} \times \vec{b} & \vec{b} \times \vec{c} & \vec{c} \times \vec{a} \end{bmatrix} = \lambda \begin{bmatrix} \vec{a} & \vec{b} & \vec{c} \end{bmatrix}^2$$, then $$\lambda$$ is equal to:

Let us agree on the usual shorthand $$[\,\vec{u}\;\vec{v}\;\vec{w}\,]=\vec{u}\cdot(\vec{v}\times\vec{w})$$; this is the scalar triple product (a single number).

The statement in the question can therefore be rewritten as

$$[\ \vec{a}\times\vec{b}\;\;\vec{b}\times\vec{c}\;\;\vec{c}\times\vec{a}\,]=\lambda\,[\,\vec{a}\;\vec{b}\;\vec{c}\,]^2.$$

We shall evaluate the left-hand scalar triple product step by step.

First put

$$\vec{p}=\vec{a}\times\vec{b},\qquad \vec{q}=\vec{b}\times\vec{c},\qquad \vec{r}=\vec{c}\times\vec{a}.$$

Then the required quantity is simply $$[\,\vec{p}\;\vec{q}\;\vec{r}\,]= \vec{p}\cdot(\vec{q}\times\vec{r}).$$

We begin by finding $$\vec{q}\times\vec{r}=(\vec{b}\times\vec{c})\times(\vec{c}\times\vec{a}).$$

The vector identity for the cross‐product of two cross‐products is

$$(\vec{u}\times\vec{v})\times(\vec{w}\times\vec{x}) =[\,\vec{u}\;\vec{v}\;\vec{x}\,]\,\vec{w}-[\,\vec{u}\;\vec{v}\;\vec{w}\,]\,\vec{x}.$$

Choosing $$\vec{u}=\vec{b},\;\vec{v}=\vec{c},\;\vec{w}=\vec{c},\;\vec{x}=\vec{a},$$ we obtain

$$ (\vec{b}\times\vec{c})\times(\vec{c}\times\vec{a}) =[\,\vec{b}\;\vec{c}\;\vec{a}\,]\,\vec{c}-[\,\vec{b}\;\vec{c}\;\vec{c}\,]\,\vec{a}. $$

Because any scalar triple product containing two identical vectors is zero, the second term vanishes:

$$[\,\vec{b}\;\vec{c}\;\vec{c}\,]=0.$$ Hence

$$\vec{q}\times\vec{r}=(\vec{b}\times\vec{c})\times(\vec{c}\times\vec{a}) =[\,\vec{b}\;\vec{c}\;\vec{a}\,]\,\vec{c}.$$

The scalar triple product is cyclic, so $$[\,\vec{b}\;\vec{c}\;\vec{a}\,]=[\,\vec{a}\;\vec{b}\;\vec{c}\,].$$ Denote this common value by

$$S=[\,\vec{a}\;\vec{b}\;\vec{c}\,].$$

Consequently

$$\vec{q}\times\vec{r}=S\,\vec{c}.$$

Now compute $$\vec{p}\cdot(\vec{q}\times\vec{r}).$$ We have

$$\vec{p}=\vec{a}\times\vec{b},\quad \vec{q}\times\vec{r}=S\,\vec{c},$$ so

$$ \vec{p}\cdot(\vec{q}\times\vec{r}) =(\vec{a}\times\vec{b})\cdot(S\,\vec{c}) =S\,(\vec{a}\times\vec{b})\cdot\vec{c}. $$

The dot product in the last expression is again the scalar triple product of $$\vec{a},\vec{b},\vec{c}$$, namely

$$ (\vec{a}\times\vec{b})\cdot\vec{c}=[\,\vec{a}\;\vec{b}\;\vec{c}\,]=S.$$ Therefore

$$[\,\vec{p}\;\vec{q}\;\vec{r}\,]=S\cdot S=S^{2}=[\,\vec{a}\;\vec{b}\;\vec{c}\,]^{2}.$$

Returning to the original notation, the relation becomes

$$[\,\vec{a}\times\vec{b}\;\;\vec{b}\times\vec{c}\;\;\vec{c}\times\vec{a}\,] =[\,\vec{a}\;\vec{b}\;\vec{c}\,]^{2}.$$

Comparing this with the given equation $$[\,\vec{a}\times\vec{b}\;\;\vec{b}\times\vec{c}\;\;\vec{c}\times\vec{a}\,] =\lambda\,[\,\vec{a}\;\vec{b}\;\vec{c}\,]^{2},$$ we see that

$$\lambda=1.$$

Hence, the correct answer is Option B.