Prove that the lines `(x+1)/3=(y+3)/5=(z+5)/7a n d(x-2)/1=(y-4)/4=(z-6)/7` are coplanar . Aslo, find the plane containing these two lines.

Explanation

Here, `(x_(1),y_(1),z_(1)) = (-1,-3,-5)`
`(x_(2),y_(2),z_(2)) = (2,4,6)`
`a_(1),b_(1),c_(1) = 3,5,7` and `a_(2),b_(2),c_(2)=1,4,7`
Now, `|{:(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)):}|`
`= |{:(3,7,11),(3,5,7),(1,4,7):}|`
`= 3(35-28)-7(21-7)+11(12-5)`
`= 21-98+77 = 0`
Therefore the given lines are coplanar.
Equation of plane containing the given lines
`|{:(x-x_(1),y-y_(1),z-z_(1)),(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)):}|`
`rArr |{:(x+1,y+3,z+5),(3,5,7),(1,4,7):}| = 0`
`rArr (x+1)(35-28)-(y+3)(21-7)+(z+5)(12-5)=0`
`rArr 7(z+1)-14(y+3)+7(x+5)=0`
`rArr (x+1)-2(y+3)+(z+5)=0`
`rArr x-2y+z=0`