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 $\frac{x}{4}=\frac{y-2}{2}=\frac{z+3}{3}$ is:

Step 1: Define the lines and point

We are given two lines, $L_1$ and $L_2$, and a point $P_0$. We need to find the distance from $P_0$ to $L_1$ measured along $L_2$. First, we write the parametric equations for both lines. Let $L_1$ be parameterized by $\lambda$ and $L_2$ by $\mu$.

Step 2: Find a general point on L2

Any point on line $L_2$ can be represented by the coordinates $(4\mu, 2\mu+2, 3\mu-3)$ by setting each part of the equation for $L_2$ equal to $\mu$. This point will be used to form a line segment from $P_0$ to $L_1$.

Step 3: Find the equation of the line segment from P0 to L1

The line segment connecting $P_0(1{,}4{,}0)$ to a general point $P_2(4\mu, 2\mu+2, 3\mu-3)$ on $L_2$ has direction vector $\vec{P_0P_2}$. This vector represents the line along which we are measuring the distance.

Step 4: Determine the intersection point on L1

The line segment $\vec{P_0P_2}$ must intersect $L_1$. This means the point $P_2$ must lie on $L_1$. We substitute the coordinates of $P_2$ into the equation for $L_1$ and solve for $\mu$. We equate the first two ratios to find $\mu$.

Step 5: Calculate the coordinates of P2

Now that we have the value of $\mu$, we can find the exact coordinates of the point $P_2$ on $L_2$ that lies on $L_1$. This point $P_2$ is the point on $L_1$ that is reached from $P_0$ by traveling along $L_2$.

Step 6: Calculate the distance between P0 and P2

Finally, we calculate the distance between the point $P_0(1{,}4{,}0)$ and the point $P_2\left(-\frac{7}{2}, \frac{1}{4}, -\frac{45}{8}\right)$ using the distance formula in 3D. This distance is the required distance of $L_1$ from $P_0$ along $L_2$.

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