Question

The coordinates of a point on the line $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$ at a distance of $\frac{6}{\sqrt{2}}$ from the point $(1,2,3)$ is

• A. $(56,43,111)$
• B. $\left(\frac{56}{17},\frac{43}{17},\frac{111}{17}\right)$
• C. $(2,1,3)$
• D. $(-2,-1,-3)$

Answer 1

Answer: (B)

Given equation of line is
$\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda \text{(say)}$
$\Rightarrow x=3\lambda -2$, $y=2\lambda -1,z=2\lambda +3\dots \left(1\right)$
$\therefore$ Coordinates of any point on the line are $\left(3\lambda -2,2\lambda -1,2\lambda +3\right)$
The distance between this point and $\left(1,2,3\right)$ is $\frac{6}{\sqrt{2}}$
$\therefore \sqrt{{\left(3\lambda -2-1\right)}^2+{\left(2\lambda -1-2\right)}^2+{\left(2\lambda +3-3\right)}^2}=\frac{6}{\sqrt{2}}$
$\Rightarrow 9{\lambda }^2+9-18\lambda +4{\lambda }^2+9-12\lambda +4{\lambda }^2=18$
$\Rightarrow 17{\lambda }^2-30\lambda =0$
$\Rightarrow \lambda \left(17\lambda -30\right)=0$
$\Rightarrow \lambda =0,\frac{30}{17}$
Substituting the values of $\lambda$ in $\left(1\right)$, we get the required point as $\left(-2,-1,3\right)$ or
$\left(\frac{56}{17},\frac{43}{17},\frac{111}{17}\right)$.