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) a. 1 b. 5 c. 3 d. 4

Two lines L _(1) :x =5 , (y)/(3 - alpha) = (z)/(-2) and L _(2) : x = alpa , (y)/(-1) = (z)/(2- alpha ) are coplanar. Then alpha can take value (s)

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 :

The lines (x-3)/(1)=(y-2)/(2)=(z-3)/(-l) and (x-1)/(l)=(y-2)/(2)=(z-1)/(4) are coplanar if value of l is

cosider Lines L_(1):(x-alpha)/(1)=(y)/(-2)=(z+beta)/(2)*L_(2):x=alpha,(y)/(-alpha)=(z+alpha)/(2-beta) Plane p:2x+2y+z+7=0. Let line L_(2) line in plane p, then

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