1. Key Concepts and Formulas

• Foot of the Perpendicular from a Point to a Line: If $N$ is the foot of the perpendicular from point $P$ onto line $L$, then $N$ lies on $L$, and the vector $\vec{PN}$ is perpendicular to the direction vector of $L$. This means their dot product is zero.
• Equation of a Line and Direction Ratios: A line passing through $(x_1, y_1, z_1)$ with direction ratios $(l, m, n)$ is given by $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$. The direction ratios $(l, m, n)$ define the direction vector $\vec{d} = l\hat{i} + m\hat{j} + n\hat{k}$.
• Shortest Distance Between Two Skew Lines: For two skew lines $L_1: \vec{r} = \vec{a_1} + \lambda \vec{p}$ and $L_2: \vec{r} = \vec{a_2} + \mu \vec{q}$, the shortest distance $d$ is given by the formula: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right|$ Here, $\vec{a_1}$ and $\vec{a_2}$ are position vectors of points on $L_1$ and $L_2$ respectively, and $\vec{p}$ and $\vec{q}$ are their direction vectors.

2. Step-by-Step Solution

This problem requires two main parts: first, determining the complete equation of line $L_1$ using the given foot of the perpendicular, and then calculating the shortest distance between the fully defined $L_1$ and $L_2$.

Part 1: Determining the Equation of Line $L_1$

We are given the point $P(4, 3, 8)$ and the foot of the perpendicular from $P$ onto line $L_1$ as $N(3, 5, 7)$. The equation of line $L_1$ is ${L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}$. We need to find $a, b,$ and $l$.

Step 1: Use the property that the foot of the perpendicular $N$ lies on $L_1$. Since $N(3, 5, 7)$ is a point on $L_1$, its coordinates must satisfy the equation of $L_1$. This allows us to establish relationships between $a, b,$ and $l$.

Substitute $x=3, y=5, z=7$ into the equation of $L_1$: $\frac{3 - a}{l} = \frac{5 - 2}{3} = \frac{7 - b}{4}$ Simplify the middle term: $\frac{3 - a}{l} = 1 = \frac{7 - b}{4}$ From this, we equate the first and third parts to 1:

1. $\frac{3 - a}{l} = 1 \implies 3 - a = l \implies a = 3 - l \quad \text{(Equation A)}$
2. $\frac{7 - b}{4} = 1 \implies 7 - b = 4 \implies b = 3 \quad \text{(Value of } b \text{ determined)}$

Step 2: Use the property that $\vec{PN}$ is perpendicular to $L_1$. The line segment $PN$ is perpendicular to $L_1$. This means the dot product of the direction ratios of $\vec{PN}$ and the direction ratios of $L_1$ must be zero.

First, find the direction ratios (DRs) of the vector $\vec{PN}$: $P(4, 3, 8)$ and $N(3, 5, 7)$. $DRs(\vec{PN}) = (x_N - x_P, y_N - y_P, z_N - z_P) = (3 - 4, 5 - 3, 7 - 8) = (-1, 2, -1)$.

Next, identify the direction ratios of line $L_1$ from its equation: $DRs(L_1) = (l, 3, 4)$.

Since $\vec{PN}$ is perpendicular to $L_1$, their dot product is zero: $(-1)(l) + (2)(3) + (-1)(4) = 0$ $-l + 6 - 4 = 0$ $-l + 2 = 0$ $l = 2 \quad \text{(Value of } l \text{ determined)}$

Now, substitute the value of $l=2$ into Equation A to find $a$: $a = 3 - l = 3 - 2 = 1 \quad \text{(Value of } a \text{ determined)}$

With $a=1, b=3,$ and $l=2$, the complete equation of line $L_1$ is: ${L_1}: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$

Part 2: Calculating the Shortest Distance Between $L_1$ and $L_2$

Now that $L_1$ is fully defined, we will calculate the shortest distance between $L_1$ and $L_2$. The lines are: $L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ $L_2: \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$

Step 3: Extract necessary vectors for the shortest distance formula. We convert the Cartesian equations of the lines into vector form $\vec{r} = \vec{a} + \lambda \vec{d}$.

For $L_1$: A point on $L_1$ is $(1, 2, 3)$, so $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$. The direction vector of $L_1$ is $\vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

For $L_2$: A point on $L_2$ is $(2, 4, 5)$, so $\vec{a_2} = 2\hat{i} + 4\hat{j} + 5\hat{k}$. The direction vector of $L_2$ is $\vec{q} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.

Step 4: Calculate the vector $(\vec{a_2} - \vec{a_1})$. This vector connects a point on $L_1$ to a point on $L_2$. $\vec{a_2} - \vec{a_1} = (2\hat{i} + 4\hat{j} + 5\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k})$ $\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$

Step 5: Calculate the cross product $(\vec{p} \times \vec{q})$. This vector is perpendicular to both $L_1$ and $L_2$, representing the direction of the shortest distance. $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$ $= \hat{i}(3 \cdot 5 - 4 \cdot 4) - \hat{j}(2 \cdot 5 - 4 \cdot 3) + \hat{k}(2 \cdot 4 - 3 \cdot 3)$ $= \hat{i}(15 - 16) - \hat{j}(10 - 12) + \hat{k}(8 - 9)$ $= -\hat{i} + 2\hat{j} - \hat{k}$

Step 6: Calculate the magnitude of $(\vec{p} \times \vec{q})$. This is the denominator of the shortest distance formula. $|\vec{p} \times \vec{q}| = \sqrt{(-1)^2 + (2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

Step 7: Calculate the scalar triple product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} \times \vec{q})$. This is the numerator (before taking absolute value) of the shortest distance formula. $(\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (-\hat{i} + 2\hat{j} - \hat{k})$ $= (1)(-1) + (2)(2) + (2)(-1)$ $= -1 + 4 - 2 = 1$

Step 8: Apply the shortest distance formula. $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right| = \left| \frac{1}{\sqrt{6}} \right| = \frac{1}{\sqrt{6}}$

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs when calculating direction ratios, cross products, and dot products. A single sign error can propagate and lead to an incorrect final answer.
• Correctly Identifying Vectors: Ensure you correctly extract the point vectors ($\vec{a_1}, \vec{a_2}$) and direction vectors ($\vec{p}, \vec{q}$) from the line equations. Remember that $(x-x_1)/l$ implies the line passes through $(x_1, y_1, z_1)$ and has direction ratios $(l, m, n)$.
• Perpendicularity Condition: Always remember that the dot product of the direction vectors of two perpendicular lines (or a line and a segment perpendicular to it) must be zero. This is a fundamental concept in 3D geometry.

4. Summary

This problem is a comprehensive application of 3D geometry principles. It first involves utilizing the properties of the foot of the perpendicular from a point to a line to fully define the unknown parameters of line $L_1$. This process uses both the condition that the foot lies on the line and that the segment connecting the point to the foot is perpendicular to the line. Once $L_1$ is determined, the problem transitions to finding the shortest distance between two skew lines, which involves converting the line equations to vector form, performing vector operations (subtraction, cross product, dot product, magnitude), and applying the standard shortest distance formula.

The final answer is $\boxed{\frac{1}{\sqrt{6}}}$, which corresponds to option (A).