If $[a\times bb\times cc\times a]=\lambda {[abc]}^2$, then $\lambda$is equal to

The correct option is B

$1$

Find the value of $\mathit{\lambda }$:

Given that $[a\times bb\times cc\times a]=\lambda {[abc]}^2$

Taking LHS, $[a\times bb\times cc\times a]$

$=(a\times b).\left(\right(b\times c)\times (c\times a\left)\right)$

So, $(a\times b)\times (c\times d)=[abd]c-[abc]d$

Thus, $(a\times b)([bca]c-[bcc]a$

Since,$[bcc]=0$

Therefore $(a\times b)([bca\left]c\right)$

$$\begin{gathered} =(a\times b).c[bca] \\ =[abc][bca] \\ ={[abc]}^2\mathbf{\left[}\mathbf{\because }\mathbf{\left[}\mathbf{b}\mathbf{c}\mathbf{a}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\mathbf{a}\mathbf{b}\mathbf{c}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\mathbf{c}\mathbf{a}\mathbf{b}\mathbf{\right]}\mathbf{\right]}\mathbf{} \end{gathered}$$

As per the given question,

$\lambda {[abc]}^2={[abc]}^2$

Thus, $\lambda =1$

Hence, the correct option is (B).