Equation of the tangent to the curve y(x–2) (x–3)–x + 7 = 0 where it crosses the x-axis is

The given curve is $y=\frac{x-7}{\left(x-2\right)\left(x-3\right)}$

Differentiate both sides 

$\frac{dy}{dx}=\frac{x^2-5x+6-\left(x-7\right)\left(2x-5\right)}{{\left(x^2-5x+6\right)}^2}$

The slope of the tangent to the curve where the curve crosses the $x-$axis, $\left(7,0\right)$ is $\begin{array}{rcl}m& =& \frac{1}{20}\end{array}$

Therefore, the equation of the tangent is 

$\begin{array}{rcl}y-0& =& \frac{1}{20}\left(x-7\right)\\ 20y& =& x-7\\ x-20y+7& =& 0\end{array}$

The equation of the required line is $x-20y+7=0$