Let the lines l₁: (x+5)/3=(y+4)/1=(z-α)/-2 and l₂: 3 x+2 y+z-2=0=x-3 y+2 z-13 be coplanar. If the point P(a, b, c) on l₁ is nearest to the point Q(-4,-3,2), then |a|+|b|+|c| is equal to

Solution

Maths sir · 4 steps

Put l₁ inside a plane through l₂ to get α, then drop a perpendicular from Q onto l₁.

1. Plane through l2. Every plane containing l₂ has the form 3x+2y+z-2+λ(x-3y+2z-13)=0, with normal (3+λ, 2-3λ, 1+2λ). Since l₁ lies in this plane, its direction (3,1,-2) is perpendicular to that normal: 3(3+λ)+(2-3λ)-2(1+2λ)=9-4λ=0, so λ=9/4.
2. Find alpha. The point (-5,-4,α) of l₁ must also lie in the plane: -25+α+9/4(2α-6)=0. This gives 11/2α=77/2, so α=7.
3. Nearest point. A general point of l₁ is P(-5+3t, -4+t, 7-2t), so PQ⃗=(1-3t, 1-t, 2t-5). P is nearest to Q when PQ⃗ is perpendicular to the direction (3,1,-2).

   3(1-3t)+(1-t)-2(2t-5)=14-14t=0 ⇒ t=1

4. So. P=(-2,-3,5), and |a|+|b|+|c|=2+3+5= 10, which is option (c).

The idea to keep. When a line is given as two planes, write the plane family P₁+λ P₂=0 to handle coplanarity. The nearest point on a line is the foot of the perpendicular: set PQ⃗·d⃗=0.

Answer from the NTA final answer key. Solution by PrepWiser.