Let the system of linear equations
-x+2 y-9 z=7
-x+3 y+7 z=9
-2 x+y+5 z=8
-3 x+y+13 z=λ
has a unique solution x=α, y=β, z=γ. Then the distance of the point
(α, β, γ) from the plane 2 x-2 y+z=λ is :

Solution

Maths sir · 4 steps

Solve any three equations for the point, then let the fourth equation tell you λ.

1. Eliminate x. Subtract the first equation from the second to get y + 16z = 2. Take twice the first minus the third to get 3y - 23z = 6.
2. Find the point. Put y = 2 - 16z into 3y - 23z = 6: 6 - 71z = 6, so z = 0 and y = 2. The first equation then gives -x + 4 = 7, so x = -3. The point is (α, β, γ) = (-3, 2, 0), and it satisfies the third equation too: 6 + 2 + 0 = 8.
3. Find lambda. A unique solution means this point must satisfy the fourth equation as well. So λ = -3(-3) + 2 + 13(0) = 11, and the plane is 2x - 2y + z = 11.

   d = (|2(-3) - 2(2) + 0 - 11|)/√(2² + (-2)² + 1²) = 21/3

4. So. The distance is 21/3 = 7, which is option (b).

The idea to keep. Four equations in three unknowns: three of them fix the point, and the fourth only works for one value of the parameter. Always find the point first, then the parameter.

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