Let O be the origin, and M and $\mathrm{N}$ be the points on the lines $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$ and $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$ respectively such that $\mathrm{MN}$ is the shortest distance between the given lines. Then $\overrightarrow{O M} \cdot \overrightarrow{O N}$ is equal to _________.

<p>$$\mathrm{L}_1: \frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{3}=\lambda \quad \operatorname{drs}(4,1,3)=\mathrm{b}_1$$</p> <p>$$\begin{aligned} & \mathrm{M}(4 \lambda+5, \lambda+4,3 \lambda+5) \\ & \mathrm{L}_2: \frac{\mathrm{x}+8}{12}=\frac{\mathrm{y}+2}{5}=\frac{\mathrm{z}+11}{9}=\mu \\ & \mathrm{N}(12 \mu-8,5 \mu-2,9 \mu-11) \\ & \overrightarrow{\mathrm{MN}}=(4 \lambda-12 \mu+13, \lambda-5 \mu+6,3 \lambda-9 \mu+16) \quad \text{..... (1)} \end{aligned}$$</p> <p>Now</p> <p>$$\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 4 & 1 & 3 \\ 12 & 5 & 9 \end{array}\right|=-6 \hat{\mathrm{i}}+8 \hat{\mathrm{k}} \quad \text{.... (2)}$$</p> <p>Equation (1) and (2)</p> <p>$$\therefore \frac{4 \lambda-12 \mu+13}{-6}=\frac{\lambda-5 \mu+6}{0}=\frac{3 \lambda-9 \mu+16}{8}$$</p> <p>I and II</p> <p>$\lambda-5 \mu+6=0 \quad \text{.... (3)}$</p> <p>I and III</p> <p>$\lambda-3 \mu+4=0 \quad \text{.... (4)}$</p> <p>Solve (3) and (4) we get</p> <p>$$\begin{aligned} \lambda= & -1, \mu=1 \\ \therefore \quad & \mathrm{M}(1,3,2) \\ & \mathrm{N}(4,3,-2) \\ \therefore \quad & \overrightarrow{\mathrm{OM}} \cdot \overrightarrow{\mathrm{ON}}=4+9-4=9 \end{aligned}$$</p>