1. Key Concepts and Formulas

• Equation of a Line in 3D (Symmetric Form): A line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ can be represented as $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. Any general point on this line can be expressed using a parameter, say $p$. If a denominator is zero, like $\frac{y-y_1}{0}$, it implies $y-y_1 = 0 \cdot p$, meaning $y=y_1$.
• Direction Ratios (DRs) of a Line Segment: For two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, the direction ratios of the line segment connecting them are $(x_2-x_1, y_2-y_1, z_2-z_1)$.
• Condition for Perpendicularity: Two lines (or a line and a line segment) with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if and only if their dot product is zero: $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
• Distance Formula in 3D: The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

2. Step-by-Step Solution

Step 1: Represent the General Point on the Line We are given the line $L: \frac{x}{1} = \frac{y-1}{0} = \frac{z+1}{-1}$. To work with this line, we represent any general point on it using a parameter, say $p$. Let $\frac{x}{1} = \frac{y-1}{0} = \frac{z+1}{-1} = p$. From this, we can express the coordinates of a general point $A(x, y, z)$ on the line:

• $x = 1 \cdot p \implies x = p$
• $y - 1 = 0 \cdot p \implies y = 1$
• $z + 1 = -1 \cdot p \implies z = -p - 1$ So, any point $A$ on the line $L$ can be written as $A(p, 1, -p-1)$.

The given external point is $P(\beta, 0, \beta)$.

Step 2: Find the Foot of the Perpendicular Let $A(p, 1, -p-1)$ be the foot of the perpendicular from point $P(\beta, 0, \beta)$ to the line $L$. First, we find the direction ratios (DRs) of the line segment $PA$: $DRs_{PA} = (x_A - x_P, y_A - y_P, z_A - z_P)$ $DRs_{PA} = (p - \beta, 1 - 0, (-p-1) - \beta)$ $DRs_{PA} = (p - \beta, 1, -p - 1 - \beta)$

Next, we identify the direction ratios (DRs) of the given line $L$ from its symmetric form: $DRs_{L} = (1, 0, -1)$ (These are the denominators in the line equation).

Since the line segment $PA$ is perpendicular to the line $L$, their direction ratios must satisfy the perpendicularity condition ($a_1a_2 + b_1b_2 + c_1c_2 = 0$). $DRs_{PA} \cdot DRs_{L} = 0$ $(p - \beta)(1) + (1)(0) + (-p - 1 - \beta)(-1) = 0$ $p - \beta + 0 + (p + 1 + \beta) = 0$ $p - \beta + p + 1 + \beta = 0$ $2p + 1 = 0$ $p = -\frac{1}{2}$

Now, substitute this value of $p$ back into the coordinates of point $A$ to find the exact coordinates of the foot of the perpendicular: $A = \left(-\frac{1}{2}, 1, -\left(-\frac{1}{2}\right) - 1\right)$ $A = \left(-\frac{1}{2}, 1, \frac{1}{2} - 1\right)$ $A = \left(-\frac{1}{2}, 1, -\frac{1}{2}\right)$

Step 3: Calculate the Perpendicular Distance and Solve for $\beta$ We are given that the length of the perpendicular $PA$ is $\sqrt{\frac{3}{2}}$. Therefore, $PA^2 = \frac{3}{2}$. Using the distance formula for points $P(\beta, 0, \beta)$ and $A\left(-\frac{1}{2}, 1, -\frac{1}{2}\right)$: $PA^2 = \left(-\frac{1}{2} - \beta\right)^2 + (1 - 0)^2 + \left(-\frac{1}{2} - \beta\right)^2$ $PA^2 = \left(\beta + \frac{1}{2}\right)^2 + (1)^2 + \left(\beta + \frac{1}{2}\right)^2$ $PA^2 = 2\left(\beta + \frac{1}{2}\right)^2 + 1$

Now, equate this expression for $PA^2$ with the given value: $2\left(\beta + \frac{1}{2}\right)^2 + 1 = \frac{3}{2}$ $2\left(\beta + \frac{1}{2}\right)^2 = \frac{3}{2} - 1$ $2\left(\beta + \frac{1}{2}\right)^2 = \frac{1}{2}$ $\left(\beta + \frac{1}{2}\right)^2 = \frac{1}{4}$

Take the square root of both sides: $\beta + \frac{1}{2} = \pm\sqrt{\frac{1}{4}}$ $\beta + \frac{1}{2} = \pm\frac{1}{2}$

This gives two possible cases for $\beta$: Case 1: $\beta + \frac{1}{2} = \frac{1}{2}$ $\beta = \frac{1}{2} - \frac{1}{2}$ $\beta = 0$

Case 2: $\beta + \frac{1}{2} = -\frac{1}{2}$ $\beta = -\frac{1}{2} - \frac{1}{2}$ $\beta = -1$

The problem statement specifies that $\beta \ne 0$. Therefore, we must choose the value $\beta = -1$.

3. Common Mistakes & Tips

• Interpreting Zero Denominators: A common mistake is to misinterpret a zero in the denominator of a line's symmetric form (e.g., $\frac{y-1}{0}$). It means the numerator must be zero for any real $p$, implying that coordinate is constant ($y=1$ in this case).
• Sign Errors: Be meticulous with signs, especially when substituting values or squaring negative terms. For example, $(-\frac{1}{2} - \beta)^2$ is equivalent to $(\beta + \frac{1}{2})^2$.
• Algebraic Precision: Double-check each step of algebraic manipulation, from solving for the parameter $p$ to simplifying the distance equation and solving for $\beta$.
• Using Constraints: Always remember to apply any given constraints (like $\beta \ne 0$) to select the correct answer from multiple possibilities.

4. Summary

This problem effectively tests the understanding of finding the perpendicular distance from a point to a line in 3D geometry. The process involves parameterizing the line, using the perpendicularity condition (dot product of direction ratios being zero) to find the foot of the perpendicular, and then applying the distance formula. The key steps are accurate representation of a general point on the line, correct application of the dot product for perpendicularity, and careful algebraic solution. The mathematical derivation consistently leads to $\beta = -1$.

5. Final Answer

The final answer is $\boxed{-1}$, which corresponds to option (D).