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)

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 :

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

The complete set of values of alpha for which the lines (x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x-3)/(2) =(y-5)/(alpha)=(z-7)/(alpha+2) are concurrent and coplanar is

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

For real numbers alpha and beta ne 0 , if the point of intersection of the straight lines ( x - alpha)/( 1) = ( y - 1)/( 2) = ( z -1)/( 3) and ( x - 4) / ( beta) = ( y - 6)/( 3) = ( z - 7)/(3) , lies on the plane x + 2y - z = 8 then alpha - beta is equal to

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