Question

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
\(3+\frac{x-11}{(y-19)^2(z-12)^2}+\frac{y-19}{(x-11)^2(z-12)^2}+\frac{z-12}{(x-11)^2(y-19)^2}-\frac{x+y+z}{14(x-11)(y-19)(z-12)}\)
is equal to :

• A. 3
• B. 39
• C. \(-45\)
• D. 0

Hint:

The plane is \(x+y+z=42\). Put \(a=x-11,\ b=y-19,\ c=z-12\) so that \(a+b+c=0\) and use \(a^3+b^3+c^3=3abc\).

Answer 1

The Correct Option is A

The plane through \((42,0,0),(0,42,0),(0,0,42)\) is \(x+y+z=42\).
Let \(a=x-11,\ b=y-19,\ c=z-12\). Then \(a+b+c=42-42=0\).
The expression becomes \(3+\frac{a}{b^2c^2}+\frac{b}{a^2c^2}+\frac{c}{a^2b^2}-\frac{42}{14abc}\).
Combine the three fractions: \(\frac{a^3+b^3+c^3}{a^2b^2c^2}\). Since \(a+b+c=0\), \(a^3+b^3+c^3=3abc\), so this equals \(\frac{3}{abc}\).
Also \(\frac{42}{14abc}=\frac{3}{abc}\).
Expression \(=3+\frac{3}{abc}-\frac{3}{abc}=3\).
Answer: (1) 3