If a plane passes through the points (-1, k, 0),(2, k,-1),(1,1,2) and is parallel to the line (x-1)/1=(2 y+1)/2=(z+1)/-1, then the value of (k²+1)/(k-1)(k-2) is :

Solution

Maths sir · 4 steps

Both points share y = k, so AB is free of k. Cross it with the line's direction for the normal.

1. Line's direction. Make the y coefficient 1 first: (2y+1)/2 = (y + 1/2)/1. So the line is (x-1)/1 = (y+1/2)/1 = (z+1)/-1, with direction d⃗ = (1, 1, -1).
2. Two directions in plane. With A(-1, k, 0) and B(2, k, -1), AB⃗ = (3, 0, -1), and k drops out. The plane is parallel to the line, so d⃗ also lies along the plane. The normal is AB⃗ × d⃗.

   n⃗ = |î, ĵ, k̂; 3, 0, -1; 1, 1, -1| = î + 2ĵ + 3k̂

3. Plane, then C. The plane through A is (x+1) + 2(y-k) + 3z = 0. Put in C(1, 1, 2): 2 + 2(1-k) + 6 = 0, so 10 - 2k = 0 and k = 5.
4. So. (k²+1)/(k-1)(k-2) = 26/(4 × 3) = 26/12 = 13/6, which is option (c).

The idea to keep. A plane parallel to a line contains that line's direction. Cross it with a vector joining two known points to get the normal, then use the last point to find the unknown.

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