Question

If $\left(1,5,35\right),\left(7,5,5\right),\left(1,\lambda ,7\right)$ and $\left(2\lambda ,1,2\right)$ are co-planar, then the sum of all possible values of $\lambda$ is:

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

Answer 1

The correct option is D

$\frac{44}{5}$

Explanation for correct option:

Step-1: Apply condition for four points are co-planer.

Let the given points, $\left(1,5,35\right),\left(7,5,5\right),\left(1,\lambda ,7\right)$

Given four points are $\left(1,5,35\right),\left(7,5,5\right),\left(1,\lambda ,7\right)$ and $\left(2\lambda ,1,2\right)$.

We know that, four points are co-planer iff$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\\ x_4-x_1& y_4-y_1& z_4-z_1\end{array}\right|=0$

$\Rightarrow$$\left|\begin{array}{ccc}7-1& 5-5& 5-35\\ 1-1& \lambda -5& 7-35\\ 2\lambda -1& 1-5& 2-35\end{array}\right|=0$

$\Rightarrow$$\left|\begin{array}{ccc}6& 0& -30\\ 0& \lambda -5& -28\\ 2\lambda -1& -4& -33\end{array}\right|=0$

Step-2: Solve determinant by sarrus rule.

$\Rightarrow$$\left\{\left(6\times \left(-33\right)\times \left(\lambda -5\right)\right)+0+0\right\}-\left\{\left(2\lambda -1\right)\left(\lambda -5\right)\left(-30\right)+0+6\times \left(-4\right)\times \left(-28\right)\right\}=0$

$\Rightarrow$ $\left\{\left(-198\lambda +990\right)\right\}-\left\{\left(2{\lambda }^2-10\lambda -\lambda +5\right)\left(-30\right)+0+672\right\}=0$

$\Rightarrow$ $\left\{\left(-198\lambda +990\right)\right\}-\left\{\left(-60{\lambda }^2+330\lambda +150\right)+672\right\}=0$

$\Rightarrow$ $60{\lambda }^2-528\lambda +468=0$

$\Rightarrow$ $10{\lambda }^2-88\lambda +78=0$

Step-3: Use formula for sum of roots of quadratic equation $-\frac{b}{a}$.

$\Rightarrow$$10{\lambda }^2-88\lambda +78=0$

$\Rightarrow$Sum of all value of $\lambda$ is $-\frac{\left(-88\right)}{10}=\frac{44}{5}$

Hence, correct answer is option $\left(D\right).$