The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is :

<p>To find the distance from the line </p> <p>$ \frac{x-2}{2} = \frac{y-6}{3} = \frac{z-3}{4} $</p> <p>to the point $(1,4,0)$ along the line </p> <p>$ \frac{x}{1} = \frac{y-2}{2} = \frac{z+3}{3}, $</p> <p>we first consider a parallel line passing through the point $(1,4,0)$. The equation of this parallel line is:</p> <p>$ \frac{x-1}{1} = \frac{y-4}{2} = \frac{z-0}{3}. $</p> <p>Next, we find the point of intersection between the given line and the parallel line, which can be expressed in parameter form as:</p> <p>$ (\lambda + 1, 2\lambda + 4, 3\lambda) = (2t + 2, 3t + 6, 4t + 3). $</p> <p>Solving for $\lambda$ in terms of $t$, from the first coordinate:</p> <p>$ \lambda = 2t + 1. $</p> <p>Using the second coordinates, we equate:</p> <p>$ 2\lambda + 4 = 3t + 6 \quad \Rightarrow \quad 2(2t + 1) + 4 = 3t + 6. $</p> <p>Simplifying gives:</p> <p>$ 4t + 2 + 4 = 3t + 6 \quad \Rightarrow \quad 4t + 6 = 3t + 6 \quad \Rightarrow \quad t = 0. $</p> <p>Substituting $t = 0$ into either line equation yields the point of intersection (POI) as:</p> <p>$ (2, 6, 3). $</p> <p>The distance from the point $(1, 4, 0)$ to the POI $(2, 6, 3)$ is computed as:</p> <p>$ \sqrt{(2-1)^2 + (6-4)^2 + (3-0)^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}. $</p>