Vector Algebra — Class 12 Maths Solution

Let $\overrightarrow {\rm{a}} = \lambda \widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}},\overrightarrow {\rm{b}}$

$= \widehat {\rm{i}} + \lambda \widehat {\rm{j}} - \widehat {\rm{k}}$ and $\overrightarrow {\rm{c}} = 2\widehat {\rm{i}} - \widehat {\rm{j}} + \lambda \widehat {\rm{k}}$

For $\overrightarrow {\rm{a}} ,\overrightarrow {\rm{b}}$ and $\overrightarrow {\rm{c}}$ to be coplanar,

$\left| {\begin{array}{cccccccccccccccccccc}\lambda &1&2\\1&\lambda &{ - 1}\\2&{ - 1}&\lambda \end{array}} \right| = 0$

$\Rightarrow$ $\lambda \left( {{\lambda ^2} - 1} \right) - 1(\lambda + 2) + 2( - 1 - 2\lambda ) = 0$

$\Rightarrow$ ${\lambda ^3} - \lambda - \lambda - 2 - 2 - 4\lambda = 0$
$\Rightarrow$ ${\lambda ^3} - 6\lambda - 4 = 0$

$\Rightarrow$ $(\lambda + 2)\left( {{\lambda ^2} - 2\lambda - 2} \right) = 0$

$\Rightarrow$ $\lambda = - 2$ or $\lambda = \frac{{2 \pm \sqrt {12} }}{2}$

$\Rightarrow$ $\lambda = - 2$ or $\lambda = \frac{{2 \pm 2\sqrt 3 }}{2} = 1 \pm \sqrt 3$