If the line x/3 = (y+2)/-1 = (z+λ)/-2 lies on the plane 2x+4y+3z=2, then the shortest distance between this line and the line x/12 = y/9 = z/4 is

The correct option is A
0
x/3 = (y+2)/-1 = (z+λ)/-2 = k
Let P be any point on the line, P = (3k, -k-2, -2k-λ)
P lies on the plane 2x+4y+3z=2
2(3k) + 4(-k-2) + 3(-2k-λ) = 2
6k - 4k - 8 - 6k - 3λ = 2
-4k - 3λ = 10
Another line is x/12 = y/9 = z/4 (line2)
Shortest distance be d.
d = √(((x1-x2)^2+(y1-y2)^2+(z1-z2)^2)/(a1^2+b1^2+c1^2))
where a1, b1, c1 are d.r's of lines
→ √(3^2+(-1)^2+(-2)^2)/√(12^2+9^2+4^2) = √(14)/√(225) = √14/15
If the line lies on the plane, then any point on the line must satisfy the equation of the plane.
2(3k) + 4(-k-2) + 3(-2k-λ) = 2
6k - 4k - 8 - 6k - 3λ = 2
-4k - 3λ = 10
If λ = -10/3 - 4k/3
Let the line be x/3 = (y+2)/-1 = (z+λ)/-2 = k
Then x = 3k, y = -k-2, z = -2k-λ
Substituting in 2x+4y+3z=2:
2(3k) + 4(-k-2) + 3(-2k-λ) = 2
6k - 4k - 8 - 6k - 3λ = 2
-4k - 3λ = 10
For the lines to intersect, the shortest distance must be 0.
The direction ratios of the first line are (3, -1, -2)
The direction ratios of the second line are (12, 9, 4)
Since the lines are not parallel, they must intersect for the shortest distance to be 0.
Let's check if they intersect:
3k = 12m, -k-2 = 9m, -2k-λ = 4m
From the first two equations, k = 4m, -4m - 2 = 9m, 13m = -2, m = -2/13, k = -8/13
Substituting in the third equation:
-2(-8/13) - λ = 4(-2/13)
16/13 - λ = -8/13
λ = 24/13
Since we found a value of λ, the lines intersect, and the shortest distance is 0.