Let

$L_1:\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}=\lambda$

Point on $L_1\equiv \left(\lambda +1,2\lambda +2,\lambda -3\right)$

And,

$L_2:\frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1}=\mu$

Point on $L_2\equiv \left(2\mu +a,3\mu -2,\mu +3\right)$

For finding $P$, we must have

$\lambda -3=\mu +3\Rightarrow \lambda =\mu +6 ...\left(\mathrm{i}\right)$

$2\lambda +2=3\mu -2\Rightarrow 2\lambda =3\mu -4 ...\left(\mathrm{ii}\right)$

Solving $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$, we get

$\lambda =22$ and $\mu =16$

Therefore,

$P\equiv \left(23,46,19\right)$

And,

$2\mu +a=\lambda +1$

$\Rightarrow 32+a=22+1$

$\Rightarrow a=-9$

Distance of point $P\left(23,46,19\right)$ from the plane $z+9=0$ is

$=\left|\frac{19+9}{\sqrt{0^2+0^2+1^2}}\right|=28$ units