Q.Find the point on the line (x+2)/3 = (y+1)/2 = (z-3)/2 at a distance 3 sqrt(2) units from the point (1, 2, 3).

Write a general point on the line using a parameter $t$, set its squared distance from the given point equal to $18$, and solve for $t$.

Parametrize the line $\dfrac{x+2}{3}=\dfrac{y+1}{2}=\dfrac{z-3}{2}=t$:

$x=-2+3t,\quad y=-1+2t,\quad z=3+2t$

Distance from $(1,2,3)$:

$x-1=3t-3=3(t-1),\quad y-2=2t-3,\quad z-3=2t$

Squared distance:

$9(t-1)^2+(2t-3)^2+(2t)^2=9(t^2-2t+1)+(4t^2-12t+9)+4t^2=17t^2-30t+18$

Set equal to $(3\sqrt2)^2=18$:

$17t^2-30t+18=18 \;\Rightarrow\; 17t^2-30t=0 \;\Rightarrow\; t(17t-30)=0$

$t=0 \quad\text{or}\quad t=\dfrac{30}{17}$

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