A plane P contains the line of intersection of the plane r⃗ ·(î+ĵ+k̂)=6 and r⃗ ·(2 î+3 ĵ+4 k̂)=-5. If P passes through the point (0,2,-2), then the square of distance of the point (12,12,18) from the plane P is :

Solution

Maths sir · 4 steps

Any plane through the common line is P₁ + λ P₂ = 0; the given point fixes λ.

1. Family of planes. In Cartesian form the planes are x + y + z - 6 = 0 and 2x + 3y + 4z + 5 = 0. Every plane through their line of intersection has the form (x + y + z - 6) + λ(2x + 3y + 4z + 5) = 0.
2. Find lambda. Put in (0, 2, -2): (0 + 2 - 2 - 6) + λ(0 + 6 - 8 + 5) = -6 + 3λ = 0, so λ = 2. The plane is 5x + 7y + 9z + 4 = 0.

   d² = (5(12) + 7(12) + 9(18) + 4)²/(5² + 7² + 9²) = 310²/155

3. Simplify. The numerator is 60 + 84 + 162 + 4 = 310. Since 310 = 2 × 155, d² = (310 × 310)/155 = 2 × 310 = 620.
4. So. The square of the distance is 620, which is option (b).

The idea to keep. For a plane through the line where two planes meet, write P₁ + λ P₂ = 0 and let the extra condition fix λ. When the distance is asked squared, square the numerator before dividing.

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