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. $3x^2-10x+21=0$
• B. $3x^2+10x-13=0$
• C. $3x^2-10x+7=0$
• D. $3x^2+10x+7=0$

Answer 1

The correct option is C $3x^2-10x+7=0$

$\left|\frac{3\cdot 1+4\cdot 1-12\cdot \lambda +13}{\sqrt{3^2+4^2+(-12{)}^2}}\right|=\left|\frac{3(-3)-12\cdot 1+13}{\sqrt{3^2+4^2+(-12{)}^2}}\right|$
$\Rightarrow |20-12\lambda |=8$
$\Rightarrow 20-12\lambda =\pm 8$
$\Rightarrow \lambda =1,\frac{7}{3}$
$\Rightarrow \lambda$ satisfies the equation $x^2-(1+\frac{7}{3})x+\frac{7}{3}=0$
$\text{i.e.},3x^2-10x+7=0$