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. 1,4,5
B. 1,2,5
C. 3,4,5
D. 2,4,5

Answer

A

Explanation

The equations of given lines are:
`L_(1):(x-5)/0=y/(3-alpha)=z/(-2),L_(2):(x-alpha)/0=y/(-1)=z/(2-alpha)`
These lines are coplanar. Therefore
`|(5-alpha, 0-0,0-0),(0,3-alpha,-2),(0,-1,2-alpha)|=0`
`implies(5-alpha){(3-alpha)(2-alpha)-2}=0`
`=((5-alpha)(6-5alpha+alpha^(2)-2)=0`
`implies(5-alpha)(alpha^(2)-5alpha+4)=0`
`implies(alpha-1)(alpha-4)(alpha-5)=0impliesalpha=1,4,5`