\( \left. \begin{array} { l } { d _ { 1 } : 2 x - 4 y + 5 = 0 \quad d _ { 1 } / l d _ { 2 } } \\ { d _ { 2 } : 3 x + a y - 4 = 0 } \end{array} \right. \quad \Rightarrow a = ? \)

To find the value of \( a \) such that the two lines defined by the equations \( d_1: 2x - 4y + 5 = 0 \) and \( d_2: 3x + ay - 4 = 0 \) are parallel, we need to set the coefficients of \( x \) and \( y \) in \( d_1 \) and \( d_2 \) proportional. First, we rewrite \( d_1 \) in the slope-intercept form (y = mx + b): \[ -4y = -2x - 5 \] \[ y = \frac{1}{2}x + \frac{5}{4} \] Here, the slope (\( m_1 \)) is \( \frac{1}{2} \). For \( d_2 \): \[ ay = -3x + 4 \] \[ y = -\frac{3}{a}x + \frac{4}{a} \] Here, the slope (\( m_2 \)) is \( -\frac{3}{a} \). Since the lines are parallel, their slopes must be equal: \[ \frac{1}{2} = -\frac{3}{a} \] Cross-multiplying gives: \[ a = -6 \] Thus, the value of \( a \) is \( -6 \).