Find the shortest distance between the lines `(x+2)/(-4) = (y)/(1) = (z-7)/(1)` and `(x+3)/(-4) = (y-6)/(3) = 9/2`

Explanation

Here `x_(1) = -2, y_(1) = 0, z_(1) = 7, a_(1) = -4`,
`b_(1) = 1, c_(1) = 1 x_(2) = x_(-3), y_(2) = 6, z_(2) = 0, a_(2) = -4, b_(2) = 3, c_(2) = 2`
Now, `D = (a_(1)b_(2) - a_(2)b_(1))^(2)+ (b_(1)c_(2)b_(2)c_(1))^(2) + (a_(2)c_(1)-a_(1)c_(2))^(2)`
`= (-12+4)^(2) + (2-3)^(2) + (-4+8)^(2)`
`= 64 + 1 + 16 = 81`
and shortest distance
`=(1)/(sqrt(D)) = |{:(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)):}|`
`= (1)/(sqrt(81))|{:(-1,6,-7),(-4,1,1),(-4,3,2):}|`
`=1/9[-1(2-3)-6(-8+
4)-7(-12+4)]`
`= 1/9[1+24+56] = (81)/(9) = 9` units.