If the line $3x + 4y + k = 0$ touches the ellipse $9x^2 + 16y^2 = 144$, then the value of k is.

The equation of the ellipse is $9x^2 + 16y^2 = 144$.

Dividing by $144$, we get $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$.

Here, $a^2 = 16$ and $b^2 = 9$.

The equation of the line is $3x + 4y + k = 0$, which can be written as $y = -\dfrac{3}{4}x - \dfrac{k}{4}$.

Comparing with $y = mx + c$, we get $m = -\dfrac{3}{4}$ and $c = -\dfrac{k}{4}$.

The condition for the line $y = mx + c$ to touch the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$.

Substituting the values, we get:

$\left(-\dfrac{k}{4}\right)^2 = 16\left(-\dfrac{3}{4}\right)^2 + 9$

$\dfrac{k^2}{16} = 16\left(\dfrac{9}{16}\right) + 9$

$\dfrac{k^2}{16} = 9 + 9 = 18$

$k^2 = 18 \times 16 = 288$

$k = \pm 12\sqrt{2}$

Answer: $\mp 12\sqrt{2}$