Let α x+β y+γ z=1 be the equation of a plane passing through the point (3,-2,5) and perpendicular to the line joining the points (1,2,3) and (-2,3,5). Then the value of α β y is equal to _____________.

Solution

Maths sir · 4 steps

The line gives you the normal. Write the plane, then scale it until the right side is exactly 1.

1. Find the normal. The plane is perpendicular to the line, so the line's direction is the plane's normal. Subtract the two points: (-2-1, 3-2, 5-3) = (-3, 1, 2).
2. Write the plane. Use the normal (-3, 1, 2) and the point (3, -2, 5) in the point-normal form.

   -3(x-3) + (y+2) + 2(z-5) = 0 ⇒ -3x + y + 2z = -1

3. Match the form. The given form has +1 on the right side, so multiply through by -1: 3x - y - 2z = 1. This fixes α = 3, β = -1, γ = -2.
4. So. The product asked for is αβγ (printed as αβ y): 3 × (-1) × (-2) = 6.

The common slip: -6. Reading the coefficients straight from the normal (-3, 1, 2), without rescaling so the right side is +1, gives -6 instead of 6.

The idea to keep. A plane's normal can be scaled freely, but once the equation is fixed as α x+β y+γ z=1, the coefficients are unique. Always rescale to match the given form before reading them off.

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