1. **State the problem:** We have the curve $y = x \ln x$ and a line $2x - 2y + 3 = 0$. We want to find the equation of the perpendicular to the curve that is parallel to this line. 2. **Find the slope of the given line:** Rewrite the line as $2x - 2y + 3 = 0 \Rightarrow 2y = 2x + 3 \Rightarrow y = x + \frac{3}{2}$. The slope is $m = 1$. 3. **Find the slope of the tangent to the curve:** Differentiate $y = x \ln x$ using the product rule: $$\frac{dy}{dx} = \ln x + 1$$ 4. **Slope of the perpendicular to the curve:** The slope of the perpendicular is the negative reciprocal of the tangent slope: $$m_\perp = -\frac{1}{\ln x + 1}$$ 5. **Condition for the perpendicular to be parallel to the line:** Since the line has slope 1, the perpendicular must have slope 1: $$-\frac{1}{\ln x + 1} = 1$$ 6. **Solve for $x$:** $$-1 = \ln x + 1 \Rightarrow \ln x = -2 \Rightarrow x = e^{-2}$$ 7. **Find the corresponding $y$ on the curve:** $$y = x \ln x = e^{-2} \times (-2) = -2 e^{-2}$$ 8. **Equation of the perpendicular line:** It has slope 1 and passes through $(e^{-2}, -2 e^{-2})$: $$y - (-2 e^{-2}) = 1 (x - e^{-2})$$ $$y + 2 e^{-2} = x - e^{-2}$$ $$x - y = 3 e^{-2}$$ **Final answer:** The perpendicular equation is $x - y = 3 e^{-2}$, which corresponds to option (a).