Let $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{6}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?

Step 1: Identify points and direction vectors

First, we identify a point on each line and their respective direction vectors. For line $L_1$, a point is $A_1(1, 2, 3)$ and its direction vector is $\vec{b}_{1} = (2, 3, 4)$. For line $L_2$, a point is $A_2(2, 4, 5)$ and its direction vector is $\vec{b}_{2} = (3, 4, 6)$.

Step 2: Calculate the vector connecting the points

Next, we find the vector connecting a point on $L_1$ to a point on $L_2$. This vector is $\vec{A_1A_2} = \vec{a}_{2} - \vec{a}_{1}$.

Step 3: Calculate the cross product of direction vectors

The shortest distance between two skew lines is perpendicular to both direction vectors. We find a vector normal to both $\vec{b}_{1}$ and $\vec{b}_{2}$ by calculating their cross product, $\vec{b}_{1} \times \vec{b}_{2}$.

Step 4: Calculate the shortest distance

We calculate the shortest distance between the lines using the formula for skew lines. Since the shortest distance is 0, the lines are intersecting. The line of shortest distance in this case is simply the point of intersection.

Step 5: Find the point of intersection

To find the point of intersection, we equate the general points on $L_1$ and $L_2$. This gives us a system of three linear equations. Solving the first two equations: $2\lambda - 3\mu = 1$ and $3\lambda - 4\mu = 2$. Multiplying the first by 3 and the second by 2 gives $6\lambda - 9\mu = 3$ and $6\lambda - 8\mu = 4$. Subtracting these yields $-\mu = -1$, so $\mu = 1$. Substituting $\mu = 1$ into $2\lambda - 3\mu = 1$ gives $2\lambda - 3 = 1$, so $2\lambda = 4$, and $\lambda = 2$. We verify with the third equation: $3 + 4(2) = 11$ and $5 + 6(1) = 11$. Both sides are equal, so the values are consistent.

Step 6: Determine the intersection point

Substitute $\lambda = 2$ into the equation for $L_1$ to find the coordinates of the intersection point. Alternatively, substituting $\mu = 1$ into the equation for $L_2$ would yield the same point. The point of intersection is $(5, 8, 11)$. This point lies on the line of shortest distance (which is just the point itself since the lines intersect).

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