Question

If the points $(1,1,\lambda )$ and $(-3,0,1)$ are equidistant from the plane, $3x+4y-12z+13=0$, then $\lambda$ satisfies the equation :

• A. $3x^2+10x+7=0$
• B. $3x^2+7x-7=0$
• C. $3x^2-10x+7=0$
• D. $3x^2+10x-7=0$

Answer 1

Answer: (C)

$3x^2-10x+7=0$

Distance of point $(-3,0,1)$ from plane $p_1\equiv 3x+4y-12z+13=0$ is $d_1$

$d_1=\frac{|3\times -3+4\times 0-12\times 1+13|}{\sqrt{3^2+4^2+{12}^2}}$

$=\frac{8}{\sqrt{169}}=\frac{8}{13}$

Distance of point $(1,1,\lambda )$ from plane $p\equiv 3x+4y-12z+13=0$ is $d_2$

$\therefore d_2=\frac{|3+4-12\lambda +13|}{13}$

$=\frac{|20-12\lambda |}{13}=d_1$

When, $20-12\lambda >0$

$⟹\frac{20-12\lambda }{13}=\frac{8}{13}\Rightarrow {\lambda }_1=1$

When, $20-12\lambda <0$

$⟹\frac{12\lambda -20}{13}=\frac{8}{13}$

$⟹12\lambda =28$

$⟹{\lambda }_2=\frac{28}{12}=2.33\Rightarrow {\lambda }_2=2.33$

So, ${\lambda }_1+{\lambda }_2=-(1+2.33)=-3.33$ and ${\lambda }_1\cdot {\lambda }_2=2.33$

We know that in any quadratic equation.

$x^2+({\lambda }_1+{\lambda }_2)x+{\lambda }_1\cdot {\lambda }_2=0$

$x^2-3.33x+2.33=0$

$\Rightarrow x^2-\frac{10}{3}x+\frac{7}{3}=0$

$\Rightarrow 3x^2-10x+7=0$

Hence, $\left(C\right)$ is the correct answer.