The coordinates of a point on the hyperbola $\frac{x^2}{24}-\frac{y^2}{18}=1$ which is nearest to the line 3x + 2y + 1 = 0 are

 Point P is nearest to the given line if the tangent at P is parallel to the given line.
Now, the slope of tangent at P(x₁, y₁) is 

${\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{18y_1}{24x_1}=\frac{3}{4}\frac{y_1}{x_1}\text{ which must be equal to }-3/2\text{. }$

Therefore,

$\begin{array}{l}\frac{3}{4}\frac{y_1}{x_1}=-\frac{3}{2}\\ \text{ or }{y}_1=-2x_1----\left(1\right)\end{array}$

Also, (x₁, y₁) lies on the curve. Hence, 

$x\frac{2}{24}-\frac{y_1^2}{18}=1---\left(2\right)$

Solving (1) and (2), we get two points (6, -3) and (-6, 3) of which (6, -3) is the nearest.