Equation of first line,
$\begin{array}{l}
\frac{x-6}{1}=\frac{y-2}{-2}=\frac{z-2}{2}=k \text { (say) } \\
\therefore x=k+6, y=-2 k+2, z=2 k+2
\end{array}$
Hence, general point on the first line, $\mathrm{P} \equiv(\mathrm{k}+6,-2 \mathrm{k}+2,2 \mathrm{k}+2)$
Equation of second line,
$\begin{array}{l}
\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}}{-2}=\frac{\mathrm{Z}+1}{-2}=l \text { (say) }\\
\therefore \mathrm{x}=3 l-4, \mathrm{y}=-2 l, \mathrm{z}=-2 l-1
\end{array}$
Hence, general point on the second line, $Q \equiv(3 l-4,-2 l,-2 l-1)$
Direction ratios of $\mathrm{PQ}$ are
$3 l-4-\mathrm{k}-6,-2 l+2 \mathrm{k}-2,-2 l-1-2 \mathrm{k}-2$
i.e. $3 l-\mathrm{k}-10,-2 l+2 \mathrm{k}-2,-2 l-2 \mathrm{k}-3$
Now $|\mathrm{PQ}|$ will be the shortest distance between the two lines if PQ is perpendicular to both the lines. Hence,
$\begin{array}{l}
1(3 l-\mathrm{k}-10)+(-2) \\
(-2 l+2 \mathrm{k}-2)+2(-2 l-2 \mathrm{k}-3)=0
\end{array}$
and $3(3 l-\mathrm{k}-10)+(-2)(-2 l$
$+2 \mathrm{k}-2)+(-2)(-2 \mathrm{l}-2 \mathrm{k}-3)=0$
i.e. $3 l-9 \mathrm{k}=12$ or $l-3 \mathrm{k}=4$ ...(i)
and $17 l-3 \mathrm{k}=20$ ...(ii)
Subtracting equation (i) from (ii), we get
$16 \mathrm{l}=16$
$\therefore l=1$
Putting this value of 1 in equation (i), we get
$\begin{aligned}
-3 \mathrm{k} &=3, \therefore \mathrm{k}=-1 \\
\therefore \mathrm{P} & \equiv(-1+6,-2(-1)+2,2(-1)+2) \\
& \equiv(5,4,0)
\end{aligned}$
Similarly, $\mathrm{Q}=(-1,-2,-3)$
Hence, shortest distance, PQ,
$\begin{array}{l}
=\sqrt{(-1-5)^{2}+(-2-4)^{2}+(-3-0)^{2}} \\
=\sqrt{(-6)^{2}+(-6)^{2}+(-3)^{2}}=\sqrt{36+36+9} \\
=9 \text { units }
\end{array}$