The shortest distance between the lines `(x-3)/3=(y-8)/(-1)=(z-3)/1a n d(x+3)/(-3)=(y+7)/2=(z-6)/4` is a. `sqrt(30)` b. `2sqrt(30)` c. `5sqrt(30)` d. `3sqrt(30)`

Explanation

As we have given

`(x-3)/3=(y-8)/(-1)=(z-3)/1a n d(x+3)/(-3)=(y+7)/2=(z-6)/4`

`implies` we have given

`x_1=3,y_1=8,z_1=3` and `a_1=3,b_1=−1,c_1=1`

`x_2=−3,y_2=−7,z_2=6` and `a_2=−3,b_2=2,c_2=4`

Then the shortest distance between line is

`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)|)/(sqrt((a_1b_2-a_2b_1)^2+(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2))|`

`d=|(|(-6,-15,3),(3,-1,1),(-3,2,4)|)/(sqrt(((3)(2)-(-3)(-1))^2+((-1)(4)-(2)(1))^2+((1)(-3)-(4)(3))^2))|`

...