Given line $L\mathrm{}:\mathrm{}4x+3y=\lambda$
Circle $S\mathrm{}:\mathrm{}{x}^2+y^2-6x+4y=12$

Centre $\left(3,–2\right),$ Radius $=5$
Now, the line will be tangent to the circle
$\therefore$ $p=r$
$\left|\frac{4\left(3\right)+3\left(-2\right)-\lambda }{5}\right|=5\mathrm{}\Rightarrow \mathrm{}\lambda =31,\mathrm{}–19$
So, for only one point of intersection inequality, $4x+3y\le \lambda$ must not satisfy the centre of the circle
$\therefore \mathrm{}\lambda =-19$