Thank you for reporting, we will resolve it shortly
Q.
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
Three Dimensional Geometry
Solution:
Given equation of line is
$\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda\, \text{(say)}$
$\Rightarrow x=3\lambda-2$, $y=2\lambda-1, z=2\lambda+3\quad\ldots\left(1\right)$
$\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
$\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$.