(a) Show that the following lines do not intersect each other: (x-1)/3=(y+1)/2=(z-1)/5; (x+2)/4=(y-1)/3=(z+1)/-2

Shortest Distance $\displaystyle = \dfrac{\begin{vmatrix} -3 & 2 & -2 \\ 3 & 2 & 5 \\ 4 & 3 & -2 \end{vmatrix}}{\sqrt{(2\times-2-5\times3)^2+(3\times-2-4\times5)^2+(3\times3-2\times4)^2}}$

$\displaystyle = \dfrac{-3(-19)-2(-26)-2(1)}{\sqrt{361+676+1}}$

$\displaystyle = \dfrac{57+52-2}{\sqrt{1038}}$

$\displaystyle = \dfrac{107}{\sqrt{1038}} \neq 0$

So, the line will not intersect each other.

The given lines are

$\displaystyle \dfrac{x-0}{1/2} = \dfrac{y-0}{1/3} = \dfrac{z-0}{-1}$ and $\displaystyle \dfrac{x-0}{1/6} = \dfrac{y-0}{-1} = \dfrac{z-0}{-1/4}$

Or $\displaystyle \dfrac{x}{3} = \dfrac{y}{2} = \dfrac{z}{-6}$ and $\displaystyle \dfrac{x}{2} = \dfrac{y}{-12} = \dfrac{z}{-3}$

Let $\displaystyle \theta$ be the angle between the two lines, then.

$\displaystyle \cos\theta = \left| \dfrac{(3\times2) + (2\times-12) + (-6)(-3)}{\sqrt{9+4+36}\ \sqrt{4+144+9}} \right|$

$\displaystyle = \left| \dfrac{6-24+18}{7\times\sqrt{157}} \right|$

$\displaystyle = 0$

$\displaystyle \Rightarrow \theta = 90°$