Step 1: Find the point of reflection (intersection of the two lines).
The particle travels along $ x - 2y - 3 = 0 $ and hits $ 3x - 2y - 5 = 0 $. Subtracting: $ (3x-2y-5) - (x-2y-3) = 0 \Rightarrow 2x - 2 = 0 \Rightarrow x = 1 $. Then $ 1 - 2y - 3 = 0 \Rightarrow y = -1 $. Point of hit: $ (1, -1) $.
Step 2: Find the slope of the incoming path.
Line $ x - 2y - 3 = 0 $ has slope $ m_1 = \dfrac{1}{2} $ (write as $ y = \dfrac{x-3}{2} $).
Step 3: Determine the reflected direction.
The problem states the particle gets reflected in a perpendicular direction. This means the reflected ray is perpendicular to the incoming ray. For perpendicular lines, $ m_1 \cdot m_2 = -1 $, so $ m_2 = -\dfrac{1}{m_1} = -2 $.
Step 4: Form the equation of the reflected line.
The reflected line passes through $ (1, -1) $ with slope $ -2 $: \[ y - (-1) = -2(x - 1) \Rightarrow y + 1 = -2x + 2 \Rightarrow 2x + y - 1 = 0 \]
Step 5: Check the result.
At $ (1, -1) $: $ 2(1) + (-1) - 1 = 0 $. Slope of $ 2x + y - 1 = 0 $ is $ -2 $. Product with $ \dfrac{1}{2} $ is $ -1 $, confirming perpendicularity.
Step 6: State the answer.
\[ \boxed{2x + y - 1 = 0} \]