Question

if (a-a')^2+(b-b')^2+(c-c')^2=p
 and (ab'-a'b)^2+(bc'-b'c)^2+(ca'-c'a)^2=q
then the perpendicular distance of the line 
ax+by+cz=1,a'x+b'y+c'z=1 is

Answer 1

the plane having the line is
ax+by+cz +k(a'x+b'y+c'z)=1+k
the distance from origin
$d= \frac{|1+m|}{ \sqrt{(a+ma')^{2} +(b+mb')^{2} +(c+mc')^{2} } }$
$d^2= \alpha = \frac{1+2m+m^{2} }{(a'^2+b^2+c'^2)m^2+(2aa'+2bb'+2cc')m+(a^2+b^2+c^2)}$
take (a'^2+b^2+c'^2)=A 
(2aa'+2bb'+2cc')=B 
(a^2+b^2+c^2)=C
$\alpha = \frac{1+2m+m^2}{Am^2+Bm+C}$
$(A \alpha -1)m^2+(B\alpha -2)m+(C \alpha -1)=0$
the dicriminat is greater than zero
u can see that
$\frac{4(A+C-B)}{(4AC-B^2)} = \frac{p}{q}$
$0 \leq \alpha \leq \frac{p}{q} \\ d= \sqrt{ \frac{p}{q} }$