If L1 is the line of intersection of the planes 2x-y+3z=0, x-y+z+1=0 and L2 is the line of intersection of the planes x+2y-z=0, 3x-y+2z=0, then the distance of the origin from the plane, containing the lines L1 and L2 is :

Normal to plane=2x-y+3z=0 is 𝑛1=2î-ĵ+3k
Normal to plane=x-y+z+1=0 is 𝑛2=î-ĵ-k
Normal to plane=x+2y-z=0 is 𝑛3=î+2ĵ-k
Normal to plane=3x-y+2z=0 is 𝑛4=3î-ĵ+2k
So line of intersection of planes 1 and 2 is along 𝑛1×𝑛2=î+7ĵ-k: L1
similarly the line of intersection of planes 3 and 4 is along 𝑛3×𝑛4=3î+5ĵ-7k: L2
The normal to the plane containing L1 and L2 is along (î+7ĵ-k)×(3î+5ĵ-7k)=-42î+4ĵ-22k
By inspection L1 passes through (0,1,0) so the equation of the required plane is -42x+4y-22z+d=0, since (0,1,0) is on the plane, 4+d=0 or d=-4
Thus the equation of the plane containing L1 and L2 is -42x+4y-22z-4=0 or 21x-2y+11z+2=0
The distance from origin is |2|/√(21^2+2^2+11^2) = 2/√506
Alternatively, the normal to the plane containing L1 and L2 is along (î+7ĵ-k)×(3î+5ĵ-7k) = -42î+4ĵ-22k. This vector is parallel to 21î-2ĵ+11k. Then the equation of the plane is of the form 21x-2y+11z+d=0. Since the plane contains L1, which passes through (0,0,0), d=0. Then the equation of the plane is 21x-2y+11z=0. The distance from the origin is 0. This is incorrect.
Let's use the cross product of (î+7ĵ-k) and (3î+5ĵ-7k) = (-42, 4, -22). This is parallel to (21,-2,11). The equation of the plane is 21x-2y+11z=d. If we substitute a point on L1, say (0, 1, 0) we get -2=d. Thus the equation is 21x-2y+11z=-2. The distance from the origin is |-2|/sqrt(21^2+2^2+11^2) = 2/sqrt(506) which is not among the options.
Let's assume the normal to the plane is (21,-2,11). The equation of the plane is 21x-2y+11z = d. Since (0, 1, 0) lies on L1 and the plane, -2 = d. Therefore, the equation of the plane is 21x-2y+11z = -2. Distance from origin is |-2|/sqrt(21^2+2^2+11^2) = 2/sqrt(506) ≈ 0.088
Another approach:
The normal to the plane containing L1 and L2 is given by the cross product of the direction vectors of L1 and L2. Direction vector of L1 is (1, 7, -1) and direction vector of L2 is (3, 5, -7). The cross product is (-42, 4, -22), which simplifies to (21, -2, 11). The equation of the plane is 21x - 2y + 11z = k. Since (0, 1, 0) lies on L1, substituting this point gives k = -2. The equation of the plane is 21x - 2y + 11z = -2. Distance from origin is |-2|/sqrt(21^2 + (-2)^2 + 11^2) = 2/sqrt(506) which is not one of the given options. There seems to be an error in the problem or solution provided.