The equation of the normal to the curve $y=\frac{x-7}{(x-2)(x-3)}$at (7, 0)

$20x+y-140=0$

The correct answer is Option (4) → $20x+y-140=0$

$y=\frac{x-7}{(x-2)(x-3)}=\frac{x-7}{(x^2-5x+6)}$

$y'=\frac{1}{x^2-5x+6}-\frac{(x-7)(2x-5)}{(x-2)^2(x-3)^2}$

$y']_{x=7}=\frac{1}{(7-2)(7-3)}-0=\frac{1}{4×5}$

so slope of normal = -20

eq. of normal: $y=-20(x-7)$

$20x+y=140$

$20x+y-140=0$