Let the plane P contain the line 2x + y - z - 3 = 0 = 5x - 3y + 4z + 9 and
be parallel to the line $\frac{\mathrm{x}+2}{2}=\frac{3-\mathrm{y}}{-4}=\frac{\mathrm{z}-7}{5}$. Then the distance of the point A(8, -1, -19)
from the plane P measured parallel to the line $\frac{\mathrm{x}}{-3}=\frac{\mathrm{y}-5}{4}=\frac{2-\mathrm{z}}{-12}$ is equal to

$(2\mathrm{x}+\mathrm{y}-\mathrm{z}-3)+\mathrm{\lambda }(5\mathrm{x}-3\mathrm{y}+4\mathrm{z}+9)=0$
is parallel to
 $$\begin{gathered} \frac{\mathrm{x}+2}{2}=\frac{\mathrm{y}-3}{4}=\frac{\mathrm{z}-7}{5} \\ \therefore 2(2+5\mathrm{\lambda })+4(1-3\mathrm{\lambda })+5(-1+4\mathrm{\lambda })=0 \end{gathered}$$
$\Rightarrow$$\mathrm{\lambda }=\frac{-1}{6}$ 
Equation of plane $7\mathrm{x}+9\mathrm{y}-10\mathrm{z}-27=10\to \left(1\right)$
Equation of line through A(8, -1, -19) parallel to $\frac{\mathrm{x}}{-3}=\frac{\mathrm{y}-5}{4}=\frac{\mathrm{z}-2}{12}$
$$\begin{gathered} \frac{\mathrm{x}-8}{-3}=\frac{\mathrm{y}+1}{4}=\frac{\mathrm{z}+19}{12}=\lambda \\ B(-3\mathrm{\lambda }+8,4\mathrm{\lambda }-1,12\mathrm{\lambda }-19) \end{gathered}$$
sub in(1)
$$\begin{gathered} \begin{array}{r}\Rightarrow \mathrm{\lambda }=2 \\ \mathrm{B}(2,7,5)\\ \Rightarrow \mathrm{AB}=26\end{array} \end{gathered}$$