The foci of a hyperbola are $( \pm 2,0)$ and its eccentricity is $\frac{3}{2}$. A tangent, perpendicular to the line $2 x+3 y=6$, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the $\mathrm{x}$ - and $\mathrm{y}$-axes are $\mathrm{a}$ and $\mathrm{b}$ respectively, then $|6 a|+|5 b|$ is equal to __________

<p>Given that the foci are at $(\pm 2, 0)$, the distance between the foci is $2ae = 2\times2 = 4$. <br/><br/>So, $a = \frac{4}{3}$. </p> <p>The eccentricity of the hyperbola is given as $\frac{3}{2}$. Thus, $e = \frac{3}{2}$. </p> <p>For a hyperbola, $e^2 = 1 + \frac{b^2}{a^2}$, which gives $b^2 = a^2(e^2 - 1)$.<br/><br/> Substituting the given values, we get $b^2 = \frac{16}{9} - \frac{16}{9} = \frac{20}{9}$. </p> <p>The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, or $\frac{9x^2}{16} - \frac{9y^2}{20} = 1$.</p> <p>The slope of the tangent line, which is perpendicular to the given line $2x + 3y = 6$, is $\frac{3}{2}$.</p> <p>The equation of the tangent to a hyperbola is $y = mx \pm \sqrt{a^2 m^2 - b^2}$. Substituting the given values, we get $y = \frac{3x}{2} \pm \sqrt{\frac{16}{9}\times\frac{9}{4} - \frac{20}{9}} = \frac{3x}{2} \pm \frac{4}{3}$.</p> <p>The $x$-intercept occurs when $y = 0$, so $|a| = \frac{8}{9}$, and the $y$-intercept occurs when $x = 0$, so $|b| = \frac{4}{3}$.</p> <p>Finally, $|6a| + |5b| = 6\frac{8}{9} + 5\frac{4}{3} = \frac{48}{9} + \frac{60}{9} = \frac{108}{9} = 12$.</p>