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Step-by-Step Solution
Step 1: Express the line in a parametric form
The given line is
$$
\frac{x}{1} \;=\; \frac{y - 1}{0} \;=\; \frac{z + 1}{-1}.
$$
Let the common ratio be $p$. Then the coordinates of any point $A$ on this line can be written as:
$$
x = 1 \cdot p \;=\; p, \quad
y - 1 = 0 \cdot p \;=\; 0 \;\Longrightarrow\; y = 1, \quad
z + 1 = -1 \cdot p \;\Longrightarrow\; z = -1 - p.
$$
Therefore, a general point $A$ on the line is
$$
A(p, 1, -1 - p).
$$
Step 2: Identify the point from which perpendicular is drawn
We need the length of the perpendicular from the point
$$
P(\beta,\, 0,\, \beta)
$$
to this line to be
$$
\sqrt{\frac{3}{2}},
$$
where it is given that
$$
\beta \neq 0.
$$
Step 3: Apply the perpendicularity condition
The direction ratios of the line are
$$
\langle 1, \; 0, \; -1 \rangle
$$
(from the original form
$x/1 = (y-1)/0 = (z+1)/-1$).
For $AP$ to be perpendicular to the given line, the dot product of the direction of $AP$ with the direction of the line must be zero. The vector $AP$ is:
$$
\langle p - \beta,\; 1 - 0,\; -1 - p - \beta \rangle
= \langle p - \beta,\; 1,\; -1 - p - \beta \rangle.
$$
The direction vector of the line is
$$
\langle 1,\; 0,\; -1 \rangle.
$$
Hence, the perpendicularity condition is:
$$
(p - \beta)\cdot 1 \;+\; (1)\cdot 0 \;+\; (-1 - p - \beta)\cdot(-1)
= 0.
$$
Simplifying:
$$
p - \beta + (1 + p + \beta) = 0,
$$
$$
p - \beta + 1 + p + \beta = 0,
$$
$$
2p + 1 = 0
\;\Longrightarrow\; p = -\frac{1}{2}.
$$
Step 4: Find the coordinates of the foot of the perpendicular
Substituting $p = -\tfrac{1}{2}$ into $A(p, 1, -1 - p)$, we get
$$
A\left(-\frac{1}{2},\, 1,\, -1 - \left(-\frac{1}{2}\right)\right)
= \left(-\frac{1}{2},\, 1,\, -\frac{1}{2}\right).
$$
Step 5: Use the distance formula for $AP$
We know $AP$ (the distance between $A$ and $P$) is given to be
$$
\sqrt{\frac{3}{2}}.
$$
Using the distance formula in 3D:
$$
AP = \sqrt{(p_x - P_x)^2 + (p_y - P_y)^2 + (p_z - P_z)^2}.
$$
For $A\bigl(-\tfrac{1}{2}, 1, -\tfrac{1}{2}\bigr)$ and $P(\beta, 0, \beta)$:
$$
AP = \sqrt{\left(-\frac{1}{2} - \beta\right)^2 + (1 - 0)^2 + \left(-\frac{1}{2} - \beta\right)^2}.
$$
According to the problem:
$$
AP = \sqrt{\frac{3}{2}} \;\Longrightarrow\; (AP)^2 = \frac{3}{2}.
$$
Therefore,
$$
\left(-\frac{1}{2} - \beta\right)^2 + 1 + \left(-\frac{1}{2} - \beta\right)^2 = \frac{3}{2}.
$$
This simplifies to
$$
2\left(-\frac{1}{2} - \beta\right)^2 + 1 = \frac{3}{2},
$$
or equivalently
$$
2\left(-\frac{1}{2} - \beta\right)^2 = \frac{3}{2} - 1
= \frac{1}{2}.
$$
Hence
$$
\left(-\frac{1}{2} - \beta\right)^2 = \frac{1}{4}.
$$
Step 6: Solve for $β$ and apply the condition $β \neq 0$
From
$$
\left(-\frac{1}{2} - \beta\right)^2 = \frac{1}{4},
$$
we get
$$
-\frac{1}{2} - \beta = \frac{1}{2}
\quad \text{or} \quad
-\frac{1}{2} - \beta = -\frac{1}{2}.
$$
Solving each:
1) $-\tfrac{1}{2} - \beta = \tfrac{1}{2}$ gives $\beta = -1.$
2) $-\tfrac{1}{2} - \beta = -\tfrac{1}{2}$ gives $\beta = 0,$ but this contradicts the given condition $\beta \neq 0.$
Therefore,
$$
\beta = -1.
$$
This is the value that satisfies the problem requirements.
Final Answer
$$
\boxed{\beta = -1}
$$
