Question

If $(1,5,35),(7,5,5),(1,\lambda ,7)$ and $(2\lambda ,1,2)$ are coplanar, then the sum of all possible values of $\lambda$ is

• A. $\frac{39}{5}$
• B. $-\frac{39}{5}$
• C. $\frac{44}{5}$
• D. $-\frac{44}{5}$

Answer 1

Answer: (C)

$A(1,5,35),B(7,5,5),C(1,\lambda ,7),D(2\lambda ,1,2)$
$\overline{AB}=6\hat{i}-30\hat{k},\overline{BC}=-6\hat{i}(\lambda -5)\hat{j}+2\hat{k}$
$\overrightarrow{CD}=(2\lambda -1)\hat{i}+(1-\lambda )\hat{j}-5\hat{k}$
Points are coplanar
$\Rightarrow 0=\left|\begin{array}{ccc}6& 0& -30\\ -6& \lambda -5& 2\\ 2\lambda -1& 1-\lambda & -5\end{array}\right|$
$=6(-5\lambda +25-2+2\lambda )$
$-30\left(-6+6\lambda -\left(2{\lambda }^2-\lambda -10\lambda +5\right)\right)$
$=6(-3\lambda +23)-30\left(-2{\lambda }^2+11\lambda -5-6+6\lambda \right)$
$=6(-3\lambda +23)-30\left(-2{\lambda }^2+17\lambda -11\right)$
$=6\left(-3\lambda +23+10{\lambda }^2-85\lambda +55\right)$
$=6\left(10{\lambda }^2-88\lambda +78\right)=12\left(5{\lambda }^2-44\lambda +39\right)$
$\Rightarrow 0=12\left(5{\lambda }^2-44\lambda +39\right)$
${\lambda }_1+{\lambda }_2=\frac{44}{5}$