Question

The intercepts made on the $x,y$ and $z$ axes, by the plane which bisects the line joining the points $\left(1,2,3\right)$ and $\left(-3,4,5\right)$ at right angles, are $a,b$ and $c$ respectively, then the ordered triplet $\left(a,b,c\right)$ is

• A. $\left(\frac{-9}{2},9,9\right)$
• B. $\left(\frac{9}{2},9,9\right)$
• C. $\left(9,\frac{-9}{2},9\right)$
• D. $\left(9,\frac{9}{2},9\right)$

Answer 1

Answer: (A)

Let, equation of the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.$
Mid-point $M$ of line joining $P\left(1,2,3\right)$ and $Q\left(-3,4,5\right)$ is $\left(-1,3,4\right).$
It lies on the plane i.e.
$\frac{-1}{a}+\frac{3}{b}+\frac{4}{c}=1$
Also, $PQ$ is parallel to the normal of the plane, so,
$\frac{\frac{1}{a}}{-4}=\frac{\frac{1}{b}}{2}=\frac{\frac{1}{c}}{2}=\lambda$
$\Rightarrow \frac{1}{a}=-4\lambda ,\frac{1}{b}=2\lambda ,\frac{1}{c}=2\lambda$
$\Rightarrow 4\lambda +6\lambda +8\lambda =1$
$\Rightarrow \lambda =\frac{1}{18}$
i.e. $a=-\frac{9}{2},b=9,c=9$
Hence, ordered triplet is $\left(\frac{-9}{2},9,9\right)$