Two lines `L_1: x=5, y/(3-alpha)=z/(-2)` and `L_2: x=alpha, y/(-1)=z/(2-alpha)` are coplanar. Then `alpha` can take value (s) a. `1` b. `2` c. `3` d. `4`

Options

A. 1
B. 2
C. 3
D. 4

Explanation

Lines `(x-5)/(0)= (y-0)/(3-alpha)=(z-0)/(-2) and (x-alpha)/(0)= (y)/(-1) = (z)/(2-alpha)` will be coplanar if shortest distance is zero. Thus,
`" "|{:(5-alpha,,0,,0),(0,,3-alpha,,-2),(0,,-1,,2-alpha):}|=0`
or `" "(5-alpha)(alpha^(2)-5alpha+4)=0, alpha=1, 4, 5`
or `" "alpha=1, 4`