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

Answer 1

Answer: (A)

$\frac{|3+4-12\lambda +13|}{\sqrt{9+16+144}}=\frac{|-9+0-12+13|}{\sqrt{9+16+144}}$
$\Rightarrow |12\lambda +20|=|-8|$
$\Rightarrow |-12\lambda +20|=8$
$\Rightarrow 12\lambda -20=\pm$
$\Rightarrow 12\lambda =28$ or 12
$\Rightarrow \lambda =7/3$ or
$1\Rightarrow$ Clearly $3x^2-10x+7=0$ is satisfied by $\lambda =7/3$ and 1