Let O be the origin, and M and N be the points on the lines x - 5 4 = y - 4 1 = z - 5 3 and x + 8 12 = y + 2…

Let,

$L_1:\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}=\lambda$

$\Rightarrow M(4\lambda +5,\lambda +4,3\lambda +5)$

And $L_2:\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}=\mu$

$\Rightarrow N(12\mu -8,5\mu -2,9\mu -11)$

Now, finding $\overrightarrow{MN}=(4\lambda -12\mu +13,\lambda -5\mu +6,3\lambda -9\mu +16) ..\dots \left(1\right)$

Now finding cross product of direction ratios of line we get,

${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 1& 3\\ 12& 5& 9\end{array}\right|=-6\hat{i}+8\hat{k} \dots ...\left(2\right)$

Now, solving the equation $\left(1\right) \& \left(2\right)$ as they are parallel vectors, we get,

$\therefore \frac{4\lambda -12\mu +13}{-6}=\frac{\lambda -5\mu +6}{0}=\frac{3\lambda -9\mu +16}{8}$

Taking $\frac{4\lambda -12\mu +13}{-6}=\frac{\lambda -5\mu +6}{0}$

$\Rightarrow \lambda -5\mu +6=0 \dots \left(3\right)$

Taking $\frac{4\lambda -12\mu +13}{-6}=\frac{3\lambda -9\mu +16}{8}$

$\lambda -3\mu +4=0 \dots \left(4\right)$

Solve $\left(3\right)$ and $\left(4\right)$ we get

$\lambda =-1, \mu =1$

So, $M(1,3,2)$ and $N(4,3,-2)$

Hence, $\overrightarrow{OM}·\overrightarrow{ON}=4+9-4=9$