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 :

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)

Consider the line L 1 : (x +1)/3 = (y +2)/1 = (z+ 1)/2, L2 : (x-2)/1 = (y+2)/2 = (z-3)/3

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=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=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=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=alpha, (y)/(-1)=(z)/(2-alpha) are coplanar. Then, alpha can take value(s)

If the lines L_(1):(x-1)/(3)=(y-lambda)/(1)=(z-3)/(2) and L_(2):(x-3)/(1)=(y-1)/(2)=(z-2)/(3) are coplanar, then the equation of the plane passing through the point of intersection of L_(1)&L_(2) which is at a maximum distance from the origin is

The shortest distance between the lines L_(1) : (x+1)/3 = (y+2)/1 =(z+1)/2 L_(2) = (x-2)/1 = (y+2)/2 = (z-3)/3 is