The shortest distance between the straight lines through the points $\mathrm{A}_{1}=(6,2,2)$ and $A_{2}=(-4,0,-1)$, in the directions of $(1,-2,2)$ and $(3,-2,-2)$ is

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, $\begin{array}{l} =\sqrt{(-1-5)^{2}+(-2-4)^{2}+(-3-0)^{2}} \\ =\sqrt{(-6)^{2}+(-6)^{2}+(-3)^{2}}=\sqrt{36+36+9} \\ =9 \text { units } \end{array}$