The equation of line passing through (3,-1,2) and perpendicular to the lines
`vecr=(hati+hatj-hatk)+lamda(2hati-2hatj+hatk)`
and `vecr=(2hati+hatj-3hatk) +mu(hati-2hatj+2hatk)` is

Options

A. `(x+3)/(2)=(y+1)/(3)=(z-2)/(2)`
B. `(x-3)/(3)=(y+1)/(2)=(z-2)/(2)`
C. `(x-3)/(2)=(y+1)/(3)=(z-2)/(2)`
D. `(x-3)/(2)=(y+1)/(2)=(z-2)/(3)`

Answer

C

Explanation

Direction ratio o the line perpendiuclar to the lines `vecr=veca_(1)+lamdavecb_(1) and vecr=veca_(2)+muvecb_(2)` is `alpha(vecb_(1)xxvecb_(2))`
`therefore`Direction ratio of line perpendicular to the lines
`vecr=(hati+hatj-hatk)+lamda(2hati-2hatj+hatk)`
and `vecr=(2hati+hatj-3hatk)+mu(hati-2hatj+2hatk)` is `alpha|(hati,hatj,hatk),(2,-2,1),(1,-2,2)|`
`=alpha[(-4+2)hati-(4-1)hatj+(-4+2)hatk)]`
`=alpha[-2hati-3hatj-2hatk]`
Now, equation of line passing through (3,-1,2) and parallel to `-2hati-3hatj-2hak` is
`vecr=3hati-hatj+2hatk+beta(2hati+3hatj+2hatk)`
Hence, certesian form of the above equation is
`(x-3)/(2)=(y+1)/(3)=(z-2)/(2)`