Let P be the point of intersection of the line (x + 3)/3 = (y + 2)/1 = (1 - z)/2 and the plane x+y+z=2. If the distance of the point P from the plane 3x - 4y + 12z = 32 is q, then q and 2q are the roots of the equation :

Solution

Maths sir · 4 steps

Turn the line into one parameter, drop it into the plane, and P falls right out.

1. Parametrise the line. Set each fraction equal to t. Then x = 3t - 3, y = t - 2 and, because the numerator is 1 - z, z = 1 - 2t.
2. Meet the plane. Substitute into x + y + z = 2: (3t - 3) + (t - 2) + (1 - 2t) = 2t - 4 = 2, so t = 3. That gives P = (6, 1, -5).

   q = (|3(6) - 4(1) + 12(-5) - 32|)/√(3² + 4² + 12²) = (|-78|)/13 = 6

3. Build the quadratic. The roots are q = 6 and 2q = 12. Their sum is 18 and their product is 72, so the equation is x² - (sum)x + product = 0.
4. So. The equation is x² - 18x + 72 = 0, which is option (d).

The idea to keep. For a line meeting a plane, write the line with one parameter and substitute it into the plane. For roots α, β, the equation is x² - (α + β)x + αβ = 0, so watch the minus sign on the sum.

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