Let O be the origin, and M and N be the points on the lines (x -...

Let,

${L}_{1}: \dfrac{x - 5}{4} = \dfrac{y - 4}{1} = \dfrac{z - 5}{3} = \lambda$

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

And ${L}_{2}: \dfrac{x + 8}{12} = \dfrac{y + 2}{5} = \dfrac{z + 11}{9} = \mu$

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

Now, finding $\overset{\to}{MN} = ( 4\lambda - 12\mu + 13, \lambda - 5\mu + 6, 3\lambda - 9\mu + 16 ) . . \ldots \left(1\right)</p><p>Now finding cross product$ of direction ratios of line we get,

\overset{\to}{b}1 \times \overset{\to}{b}2 = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 4 & 1 & 3 \\ 12 & 5 & 9\end{cases}\right| = - 6\overset{^}{i} + 8\overset{^}{k} \ldots . . . \left(2\right)</p><p>Now, solving the $equation \left(1\right) \text{ and } \left(2\right)$ as they are parallel vectors, we get,

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

Taking $\dfrac{4\lambda - 12\mu + 13}{- 6} = \dfrac{\lambda - 5\mu + 6}{0}</p><p>$⇒ \lambda - 5\mu + 6 = 0 \ldots \left(3\right)$</p><p>Taking$\dfrac{4\lambda - 12\mu + 13}{- 6} = \dfrac{3\lambda - 9\mu + 16}{8}

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

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, $\overset{\to}{OM} \cdot \overset{\to}{ON} = 4 + 9 - 4 = 9$