If (x,y,z) is an arbitrary point lying on a plane P, which passes through the points (42,0,0),(0,42,0),(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:
Solutions · 1
Official solution
From intercept from, equation of plane is x + y + z = 42
\[ \Rightarrow \] (x - 11) + (y - 19) + (z - 12) = 0
let a = x - 11, b = y - 19, c = z - 12
a + b + c = 0
Now, given expression is
\[3 + {a \over {{b^2}{c^2}}} + {b \over {{a^2}{c^2}}} + {c \over {{a^2}{b^2}}} - {{42} \over {14abc}}\]
\[3 + {{{a^3} + {b^3} + {c^3} - 3abc} \over {{a^2}{b^2}{c^2}}}\]
If a + b + c = 0
\[ \Rightarrow \] a3 + b3 + c3 = 3 abc
\[ \therefore \] \[3 + {0 \over {{a^2}{b^2}{c^2}}}\]
=3