Given:

$$\begin{gathered} 2x+y+12=0 ...\left(i\right) \\ kx-3y-10=0 ...\left(ii\right) \end{gathered}$$

Since, above lines are conjugate w.r.t. to the circle then pole of one passes through another.

Let pole is $\left(h,m\right)$.

Now,

$x^2+y^2-4x+3y-1=0$

$T=0$

$\Rightarrow hx+my-4\left(\frac{x+h}{2}\right)+3\left(\frac{y+m}{2}\right)-1=0$

$\Rightarrow \left(h-2\right)x+\left(m+\frac{3}{2}\right)y+\left(-2h+\frac{3m}{2}-1\right)=0 ...\left(iii\right)$

And by comparing $\left(i\right) \& \left(iii\right)$, we get

$\frac{h-2}{2}=\frac{m+{\displaystyle \frac{3}{2}}}{1}=\frac{-2h+{\displaystyle \frac{3m}{2}}-1}{12}$

By solving, we get

$$\begin{gathered} h-2m=5 ...\left(iv\right) \\ 16h-3m=22 ...\left(v\right) \end{gathered}$$

By solving, we get

$h=1 \& m=-2$

Point $\left(h,m\right)$ must satisfy $\left(ii\right)$.

Therefore, by substituting this point in the line, we get

$k=4$.