Correct Option: (C) **Step 1: Determine the slope of the given line.** Let the given line be $L_1: 2x - 3y + 6 = 0$. To find its slope, we convert the equation to the slope-intercept form $y = mx + c$, where $m$ is the slope. $2x - 3y + 6 = 0$ $3y = 2x + 6$ $y = \frac{2}{3}x + \frac{6}{3}$ $y = \frac{2}{3}x + 2$ Thus, the slope of $L_1$, denoted as $m_1$, is $m_1 = \frac{2}{3}$. **Step 2: Determine the slope of the required line.** Let the required line be $L_2$. Since $L_2$ is perpendicular to $L_1$, the product of their slopes must be $-1$. If $m_2$ is the slope of $L_2$, then $m_1 m_2 = -1$. $m_2 = -\frac{1}{m_1}$ $m_2 = -\frac{1}{2/3}$ $m_2 = -\frac{3}{2}$ **Step 3: Use the point-slope form to find the equation of the required line.** The required line $L_2$ passes through the point $(x_1, y_1) = (2, -3)$ and has a slope $m_2 = -\frac{3}{2}$. The point-slope form of a linear equation is $y - y_1 = m(x - x_1)$. Substituting the known values: $y - (-3) = -\frac{3}{2}(x - 2)$ $y + 3 = -\frac{3}{2}(x - 2)$ **Step 4: Simplify the equation to the standard form.** To eliminate the fraction, multiply both sides by $2$: $2(y + 3) = -3(x - 2)$ $2y + 6 = -3x + 6$ Rearrange the terms to bring $x$ and $y$ terms to one side, typically in the form $Ax + By + C = 0$: $3x + 2y + 6 - 6 = 0$ $3x + 2y = 0$ Comparing this final equation with the given options, it perfectly matches option (C). Let's quickly analyze why the other options are incorrect: (A) $3x - 2y = 12$. The slope of this line is $m = -\frac{3}{-2} = \frac{3}{2}$. This line is not perpendicular to $L_1$ (slope $2/3$) as $(2/3) \times (3/2) = 1 \neq -1$. (B) $2x + 3y = -5$. The slope of this line is $m = -\frac{2}{3}$. This line is perpendicular to the required line $L_2$ (slope $-3/2$), but parallel to the given line $L_1$ (slope $2/3$), not perpendicular to it. (D) $2x - 3y = 13$. The slope of this line is $m = -\frac{2}{-3} = \frac{2}{3}$. This line is parallel to the given line $L_1$, not perpendicular to it. Hence, the correct equation for the line is $3x + 2y = 0$.