Correct Option: (A) To determine the equation of line $L$, we first need to identify a point it passes through and its slope. \textbf{Step 1: Find the point of intersection of lines $L_1$ and $L_2$.} We are given the equations of two lines: $L_1: x - y + 2 = 0 \implies x - y = -2 \quad (1)$ $L_2: 2x + y - 8 = 0 \implies 2x + y = 8 \quad (2)$ To find the point of intersection, we solve this system of linear equations. Add equation $(1)$ and equation $(2)$: $(x - y) + (2x + y) = -2 + 8$ $3x = 6$ $x = 2$ Now, substitute the value of $x = 2$ into equation $(1)$: $2 - y = -2$ $-y = -4$ $y = 4$ Thus, the point of intersection of lines $L_1$ and $L_2$ is $(2, 4)$. Line $L$ passes through this point. \textbf{Step 2: Determine the slope of line $L_3$.} The equation of line $L_3$ is $3x - 4y + 5 = 0$. To find its slope, we can rearrange the equation into the slope-intercept form $y = mx + c$, where $m$ is the slope. $4y = 3x + 5$ $y = \frac{3}{4}x + \frac{5}{4}$ The slope of line $L_3$ is $m_3 = \frac{3}{4}$. \textbf{Step 3: Determine the slope of line $L$.} Line $L$ is perpendicular to line $L_3$. For two perpendicular lines with slopes $m_a$ and $m_b$, their product is $-1$, i.e., $m_a m_b = -1$. Let the slope of line $L$ be $m_L$. $m_L \cdot m_3 = -1$ $m_L \cdot \frac{3}{4} = -1$ $m_L = -\frac{4}{3}$ \textbf{Step 4: Formulate the equation of line $L$.} Line $L$ passes through the point $(x_1, y_1) = (2, 4)$ and has a slope $m_L = -\frac{4}{3}$. Using the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$: $y - 4 = -\frac{4}{3}(x - 2)$ To eliminate the fraction, multiply both sides of the equation by $3$: $3(y - 4) = -4(x - 2)$ $3y - 12 = -4x + 8$ Now, rearrange the terms to the standard form $Ax + By + C = 0$: $4x + 3y - 12 - 8 = 0$ $4x + 3y - 20 = 0$ Thus, the equation of line $L$ is $4x + 3y - 20 = 0$. Comparing this result with the given options, option (A) matches our derived equation.