Question

Let $\vec{a},\vec{b}$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vectors of four points $A,B,C$ and $D$ be $\vec{a}-\vec{b}+\vec{c},\lambda \vec{a}-3\vec{b}+4\vec{c},-\vec{a}+2\vec{b}-3\vec{c}$ and $2\vec{a}-4\vec{b}+6\vec{c}$ respectively. If $\overrightarrow{AB},\overrightarrow{AC}$ and $\overrightarrow{AD}$ are coplanar, then $\lambda$ is equal to

Answer 1

Answer: 2

$\overline{AB}=(\lambda -1)\bar{a}-2\bar{b}+3\bar{c}$
$\overline{AC}=2\bar{a}+3\bar{b}-4\bar{c}$
$\overline{AD}=\bar{a}-3\bar{b}+5\bar{c}$
$\left|\begin{array}{ccc}\lambda -1& -2& 3\\ -2& 3& -4\\ 1& -3& 5\end{array}\right|=0$
$\Rightarrow (\lambda -1)(15-12)+2(-10+4)+3(6-3)=0$
$\Rightarrow (\lambda -1)=1\Rightarrow \lambda =2$