Let the line (x)/(1) = (6 - y)/(2) = (z + 8)/(5) intersect the...

Given,

The line $\dfrac{x}{1} = \dfrac{6 - y}{2} = \dfrac{z + 8}{5}$ intersect the lines $\dfrac{x - 5}{4} = \dfrac{y - 7}{3} = \dfrac{z + 2}{1} and \dfrac{x + 3}{6} = \dfrac{3 - y}{3} = \dfrac{z - 6}{1}$ at the points $A and B$ respectively,

Now plotting the diagram, we get,

Now solving, $\dfrac{x}{1} = \dfrac{y - 6}{- 2} = \dfrac{z + 8}{5} = \lambda and \dfrac{x - 5}{4} = \dfrac{y - 7}{3} = \dfrac{z + 2}{1} = t$ we get,

$A ≡ ( \lambda, - 2\lambda + 6, 5\lambda - 8 ) ≡ ( 4t + 5, 3t + 7, t - 2 )$

Now comparing both side and solving we get, $t = - 1, \lambda = 1$

Hence, point$A ( 1, 4, – 3 )$

Now solving, $\dfrac{x}{1} = \dfrac{y - 6}{- 2} = \dfrac{z + 8}{5} = \mu$ and $\dfrac{x + 3}{6} = \dfrac{y - 3}{- 3} = \dfrac{z - 6}{1} = k$ we get,

For point $B ≡ ( \mu, - 2\mu + 6, 5\mu - 8 ) ≡ ( 6k - 3, - 3k + 3, k + 6 )$

Now comparing both side and solving we get, $k = 1, \mu = 3$

Hence, point $B ( 3, 0, 7 )$

So, mid-point of $AB = M ( 2, 2, 2 )$

Now finding, distance of $M$ from the plane $2x – 2y + z – 14 = 0$ we get,

\left|\dfrac{4 - 4 + 2 - 14}{\sqrt{{2}^{2} + {2}^{2} + {1}^{2}}}\right| = \left|\dfrac{- 12}{3}\right| = 4