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`

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 :

If the points (alpha-x,alpha,alpha),(alpha,alpha-y,alpha) and (alpha,alpha,alpha-z) are coplanar then

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

If A=[(alpha,2),(2,alpha)] and determinant (A^3)=125, then the value of alpha is (a) +-1 (b) +-2 (c) +-3 (d) +-5

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

Let the line (x-2)/(3)=(y-1)/(-5)=(z+2)/(2) lies in the plane x+3y-alpha z +beta=0 . Then, (alpha,beta) equals

The system of equations alpha(x-1)+y+z=-1, x+alpha(y-1)+z=-1 and x+y+alpha(z-1)=-1 has no solution, if alpha is equal to