Find the points on the line $\dfrac{x + 2}{3} = \dfrac{y + 1}{2} = \dfrac{z - 3}{2}$ at a distance of $3 \sqrt{2}$ units from the point $\left(1, 2, 3\right)$.
Given equation of line is: $\;$ $\dfrac{x + 2}{3} = \dfrac{y + 1}{2} = \dfrac{z - 3}{2} = \lambda$ (say)
$\implies$ $x = 3 \lambda - 2$, $\;$ $y = 2 \lambda - 1$, $\;$ $z = 2 \lambda + 3$
$\therefore \;$ Any point on the given line is: $P \left(3 \lambda -2, 2 \lambda - 1, 2 \lambda + 3\right)$
Given: Distance of $P$ from $Q \left(1, 2, 3\right)$ is $3 \sqrt{2}$ units
i.e. $\sqrt{\left(3 \lambda - 2 - 1\right)^2 + \left(2 \lambda - 1 - 2\right)^2 + \left(2 \lambda + 3 - 3\right)^2} = 3 \sqrt{2}$
i.e. $\;$ $\left(3 \lambda - 3\right)^2 + \left(2 \lambda - 3\right)^2 + 4 \lambda^2 = 18$
i.e. $\;$ $9 \lambda^2 - 18 \lambda + 9 + 4 \lambda^2 - 12 \lambda + 9 + 4 \lambda^2 = 18$
i.e. $\;$ $17 \lambda^2 - 30 \lambda = 0$
i.e. $\;$ $\lambda \left(17 \lambda - 30\right) = 0$
$\implies$ $\lambda = 0$ $\;$ or $\;$ $\lambda = \dfrac{30}{17}$
When $\lambda = 0$, $\;$ $P = \left(-2, -1, 3\right)$
When $\lambda = \dfrac{30}{17}$, $\;$ $P = \left(\dfrac{90}{17} - 2, \; \dfrac{60}{17} - 1, \; \dfrac{60}{17} + 3\right)$ $\;$ i.e. $\;$ $P = \left(\dfrac{56}{17}, \; \dfrac{43}{17}, \; \dfrac{111}{17}\right)$