Let a → , b → and c → be three non-zero non-coplanar vectors. Let the position vectors of…

Given,

$\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ be three non-zero non-coplanar vectors,

Also the position vectors of four points $A, B, C$ and $D$ be $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}, \lambda \overrightarrow{\mathrm{a}}-3\overrightarrow{\mathrm{b}}+4\overrightarrow{\mathrm{c}}$, $-\overrightarrow{a}+2\overrightarrow{b}-3\overrightarrow{c}$ and $2\overrightarrow{a}-4\overrightarrow{b}+6\overrightarrow{c}$ respectively,

So, $\overrightarrow{AB}=(\lambda -1)\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c}$

$\overrightarrow{AC}=-2\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}$

$\overrightarrow{AD}=\overrightarrow{a}-3\overrightarrow{b}+5\overrightarrow{c}$

Now using the condition of coplanar we get,

$\left|\begin{array}{ccc}\lambda -1& -2& 3\\ -2& 3& -4\\ 1& -3& 5\end{array}\right|=0$

$\Rightarrow (\lambda -1)(15-12)+2(-10+4)+3(6-3)=0$

$\Rightarrow (\lambda -1)=1\Rightarrow \lambda =2$