Two lines `L_1x=5, y/(3-alpha)=z/(-2)a n dL_2: x=alphay/(-1)=z/(2-alpha)` are coplanar. Then `alpha` can take value (s) a. `1` b. `2` c. `3` d. `4`

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

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)

The lines (x-2)/(1)=(y-3)/(1)=(z-4)/(-k) and (x-1)/(k)=(y-4)/(2)=(z-5)/(1) are coplanar, if

The lines (x-1)/(2)=(y)/(-1)=(z)/(2) and x-y+z-2=0=lambda x+3z+5 are coplanar for lambda

If for some alpha in R, the lines L_1 : (x + 1)/(2) = (y-2)/(-1) = ( z -1)/(1) and L_2 : (x + 2)/(alpha) = (y +1)/(5 - alpha) = (z + 1)/(1) are coplanar , then the line L_2 passes through the point :

Line (x-2)/alpha=(y-2)/(-3)=(z+2)/2 lies in x+3y-2z+beta=0 then alpha+beta= ?

If A=[alpha2 2alpha]a n d|A^3|=125 , then the value of alpha is a.+-1 b. +-2 c. +-3 d. +-5