A line passing through the point P(1, 2) meets the line $x+y=7$ at the distance of 3 units from P. Then the slope of this line satisfies the equation 

An equation of line through $(1, 2)$ is $y – 2 = m(x – 1)$A point on this line is of the form $Q(x, mx – m + 2)$. This point will lie on the line $x + y = 7$ if 

$\begin{array}{l}x+mx-m+2=7\Rightarrow x=\frac{m+5}{m+1}\\ \text{ As }PQ=3,P{Q}^2=9\\ \Rightarrow (x-1{)}^2+(mx-m{)}^2=9\\ \Rightarrow (x-1{)}^{2}1+m^2=9\\ \Rightarrow 16m^2+16=9m^2+18m+9\\ \Rightarrow 7m^2-18m+7=0\end{array}$

$\Rightarrow$m satisfies the equation $7x^2-18x+7=0$