$\text{ If }\mathrm{M}\left(x_o,y_o\right)\text{ is the point on the curve }3x^2-4y^2=72\text{ which is nearest to the line }$

$3x+2y+1=0,\text{ then the value of }\left(x_o+y_o\right)\text{ is equal to }$

$\text{ Slope of the give line }=-\frac{3}{2}$

$\text{ The points on the curve at which the tangent is parallel to the given line. So, }$

$\text{ differentiating both sides with respect to }\mathrm{x}\text{ of }3x^2-4y^2=72\text{ we get }$

$\begin{array}{l}\frac{dy}{dx}=\frac{3x}{4y}=\frac{-3}{2}\left(\text{ given }\right)\\ \Rightarrow \frac{x}{y}=-2\\ \text{ Now }3{\left(\frac{x}{y}\right)}^2-4=\frac{72}{y^2}\Rightarrow y^2=9\Rightarrow y=-3,3\\ \text{ So, points are }(-6,3)\text{ and }(6,-3)\end{array}$

$\text{ Now, distance of }(-6,3)\text{ from the given line }=\left|\frac{-18+6+1}{\sqrt{13}}\right|=\frac{11}{\sqrt{13}}$

$\text{ And distance of }(6,-3)\text{ from the given line }=\left|\frac{18-6+1}{\sqrt{13}}\right|=\frac{13}{\sqrt{13}}$

$\text{ Clearly, the required point is on }(-6,3)=\left(x_0,y_o\right)\text{ (given) }$

$\begin{array}{l}\text{ So, }{x}_o=-6,y_o=3\\ \text{ Hence }\left(x_0+y_0\right)=-6+3=-3\end{array}$