$=(\bar{a} \times \bar{b}) \cdot[(\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})]$
$=(\vec{a} \times \vec{b}) \cdot[(\vec{b} \times \vec{c} \cdot \vec{a}) \vec{c}-(\vec{b} \times \vec{c} \cdot \vec{c}) \vec{a}]$
$ = (\vec a \times \vec b) \cdot [[\bar b\bar c\bar a]\vec c]\quad [\vec b \times \vec c.\vec c = 0]$
$=[\bar{a} \bar{b} \bar{c}] \cdot(\bar{a} \times \bar{b} \cdot \bar{c})=[\bar{a} \bar{b} \bar{c}]^{2}$
$[\vec{a} \times \vec{b} \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}]=[\bar{a} \bar{b} \bar{c}]^{2}$
So $\lambda=1$
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$x+y+\alpha z=2$
$3 x+y+z=4$
$x+2 z=1$
have a unique solution $\left(x^{*}, y^{*}, z^{*}\right)$. If $\left(\alpha, x^{*}\right),\left(y^{*}, \alpha\right)$ and $\left(x^{*},-y^{*}\right)$ are collinear points, then the sum of absolute values of all possible values of $\alpha$ is