Let $L_1$ and $L_2$ be two parallel lines and $L_3$ a line perpendicular to $L_1$ and $L_2$ at $H$ and $P$, respectively. Points $Q$ and $R$ lie on $L_1$ such that $QR = PR$ ($Q \ne  H$). Let  $d$ be the diameter of the circle inscribed in the triangle $PQR$. Point $T$ lies $L_2$ in the same semiplane as $Q$  with respect to line $L_3$ such that $\frac{1}{TH}= \frac{1}{d}- \frac{1}{PH}$ . Let $X$ be the intersection point of $PQ$ and $TH$. Find the locus of the points $X$ as $Q$ varies on $L_1$.
1. **Define the coordinates and distances:**
   - Let \( H(0,0) \), \( P(0,p) \), and \( R(\lambda,0) \).
   - Since \( QR = PR \), we have \( QR = \sqrt{(\lambda - (\lambda - w))^2 + 0^2} = w \), where \( w = \sqrt{\lambda^2 + p^2} \).
   - Therefore, \( Q(\lambda - w, 0) \).

2. **Calculate the diameter \( d \) of the inscribed circle in \(\triangle PQR\):**
   - The semi-perimeter \( s \) of \(\triangle PQR\) is:
     \[
     s = \frac{PQ + QR + PR}{2} = \frac{p + w + w}{2} = \frac{p + 2w}{2}
     \]
   - The area \( A \) of \(\triangle PQR\) can be found using the formula for the area of a triangle with base and height:
     \[
     A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times p \times w
     \]
   - The radius \( r \) of the inscribed circle is given by:
     \[
     r = \frac{A}{s} = \frac{\frac{1}{2}pw}{\frac{p + 2w}{2}} = \frac{pw}{p + 2w}
     \]
   - The diameter \( d \) is:
     \[
     d = 2r = \frac{2pw}{p + 2w}
     \]

3. **Calculate the coordinates of point \( T \):**
   - Given the condition \(\frac{1}{TH} = \frac{1}{d} - \frac{1}{PH}\):
     \[
     \frac{1}{TH} = \frac{p + 2w}{2pw} - \frac{1}{p} = \frac{p + 2w - 2w}{2pw} = \frac{1}{2w}
     \]
   - Therefore, \( TH = 2w \).
   - Since \( T \) lies on \( L_2 \) and in the same semiplane as \( Q \) with respect to \( L_3 \), we have \( T(-\frac{p^2}{\lambda}, p) \).

4. **Find the intersection point \( X \) of \( PQ \) and \( TH \):**
   - The equation of line \( PQ \) can be written as:
     \[
     y = \frac{p}{w} (x - \lambda + w)
     \]
   - The equation of line \( TH \) can be written as:
     \[
     y = \frac{p}{\frac{p^2}{\lambda}} (x + \frac{p^2}{\lambda}) = \frac{\lambda}{p} x + p
     \]
   - Solving these two equations simultaneously:
     \[
     \frac{p}{w} (x - \lambda + w) = \frac{\lambda}{p} x + p
     \]
     \[
     \frac{p}{w} x - \frac{p}{w} \lambda + \frac{p}{w} w = \frac{\lambda}{p} x + p
     \]
     \[
     \frac{p}{w} x - \frac{p}{w} \lambda + p = \frac{\lambda}{p} x + p
     \]
     \[
     \frac{p}{w} x - \frac{\lambda}{p} x = \frac{p}{w} \lambda
     \]
     \[
     x \left( \frac{p}{w} - \frac{\lambda}{p} \right) = \frac{p}{w} \lambda
     \]
     \[
     x = \frac{\frac{p}{w} \lambda}{\frac{p}{w} - \frac{\lambda}{p}}
     \]
     \[
     x = \frac{p \lambda}{p - \frac{\lambda w}{p}}
     \]
     \[
     x = \frac{p \lambda}{\frac{p^2 - \lambda w}{p}}
     \]
     \[
     x = \frac{p^2 \lambda}{p^2 - \lambda w}
     \]

5. **Determine the locus of \( X \):**
   - The equation \( x^2 - 3y^2 + 4py - p^2 = 0 \) represents a hyperbola.
   - The locus of points \( X \) is part of this hyperbola, with the conditions \( x < 0 \) and \( y \leq \frac{p}{3} \).

The final answer is \( \boxed{ x^2 - 3y^2 + 4py - p^2 = 0 } \) with \( x < 0 \) and \( y \leq \frac{p}{3} \).