# Solve the Following Questions

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

1. $\frac{39}{5}$

2. $-\frac{39}{5}$

3. $\frac{44}{5}$

4. $-\frac{44}{5}$

Correct Option: , 3

Solution:

$\mathrm{A}(1,5,35), \mathrm{B}(7,5,5), \mathrm{C}(1, \lambda, 7), \mathrm{D}(2 \lambda, 1,2)$

$\overline{\mathrm{AB}}=6 \hat{\mathrm{i}}-30 \hat{\mathrm{k}}, \overline{\mathrm{BC}}=-6 \hat{\mathrm{i}}(\lambda-5) \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{CD}}=(2 \lambda-1) \hat{\mathrm{i}}+(1-\lambda) \hat{\mathrm{j}}-5 \hat{\mathrm{k}}$

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

$=6(-5 \lambda+25-2+2 \lambda)$

$-30\left(-6+6 \lambda-\left(2 \lambda^{2}-\lambda-10 \lambda+5\right)\right)$

$=6(-3 \lambda+23)-30\left(-2 \lambda^{2}+11 \lambda-5-6+6 \lambda\right)$

$=6(-3 \lambda+23)-30\left(-2 \lambda^{2}+17 \lambda-11\right)$

$=6\left(-3 \lambda+23+10 \lambda^{2}-85 \lambda+55\right)$

$=6\left(10 \lambda^{2}-88 \lambda+78\right)=12\left(5 \lambda^{2}-44 \lambda+39\right)$

$\Rightarrow 0=12\left(5 \lambda^{2}-44 \lambda+39\right)$

$\lambda_{1}+\lambda_{2}=\frac{44}{5}$