If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression
3 + (x - 11)/(y - 19)²(z - 12)² + (y - 19)/(x - 11)²(z - 12)² + (z - 12)/(x - 11)²(y - 19)² - (x + y + z)/14(x - 11)(y - 19)(z - 12) is equal to :

Solution

Maths sir · 4 steps

Shift to a = x-11, b = y-19, c = z-12; on this plane a+b+c=0, and the fractions cancel out.

1. Find the plane. The intercepts on all three axes are 42, so the plane is x/42+y/42+z/42=1, that is x+y+z=42.
2. Substitute. Put a = x-11, b = y-19, c = z-12. Then a+b+c = (x+y+z) - 42 = 0. Also (x+y+z)/14 = 42/14 = 3, so the last term is 3/abc.

   3 + (a³+b³+c³)/a²b²c² - 3/abc

3. Use the identity. When a+b+c=0, we have a³+b³+c³ = 3abc. So the middle fraction becomes 3abc/a²b²c² = 3/abc, which cancels the last term exactly.
4. So. Only the leading 3 is left, and the expression equals 3, option (a).

The idea to keep. When an ugly expression has shifted variables, rename them and look for their sum. If a+b+c=0, then a³+b³+c³=3abc, and that usually collapses the expression.

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