The line 2x + 3y =6, 2x + 3y = 8 cut the X-axis at A, B respectively. A line l drawn through the point (2, 2) meets the X-axis at C in such a way that abscissae of A, B, C are in arithmetic progression. Then the equation of the line l is

$$\begin{gathered} Given line 2x + 3y =6\Rightarrow \frac{x}{3}+\frac{y}{2}=1, 2x + 3y = 8\Rightarrow \frac{x}{4}+\frac{y}{{\displaystyle \frac{8}{3}}}=1 \\ cut the X-axis at A, B respectively \\ \Rightarrow A\left(3 ,0\right) and B\left(4, 0\right) \end{gathered}$$

$$\begin{gathered} Let the line l drawn through the point (2, 2) \\ meets the X-axis at C\left(a,0\right) \end{gathered}$$

And given that abscissae of A, B, C are in arithmetic progression

$$\begin{gathered} \Rightarrow 3 , 4 ,a are in A.P \\ \Rightarrow 2\left(4\right)=3+a \\ \Rightarrow a=5 \\ \therefore C \left(5 ,0\right) \end{gathered}$$

now the line l passes through (2, 2) and (5 ,0)

$$\begin{gathered} the equation is y-2=\frac{0-2}{5-2}\left(x-2\right) \\ \Rightarrow y-2=\frac{-2}{3}\left(x-2\right) \\ \Rightarrow 3y-6=-2x+4 \\ \Rightarrow 2x+3y-10=0 \end{gathered}$$