[tex]\begin{gathered} \text{Let }x\text{ be the first odd integer, then it goes that} \\ x+2\rightarrow\text{second odd integer} \\ x+4\rightarrow\text{third odd integer} \\ \text{If 6 is subtracted from the third of three consecutive odd integer, translates to} \\ (x+4)-6\Rightarrow x+4-6\Rightarrow x-2 \\ \text{and the result is multiplied by 2, which translates to} \\ (x-2)\cdot2\Rightarrow2x-4 \\ \text{the answer is 23 less than the sum of the first and twice of the second integers} \\ (x)+2(x+2)\rightarrow x+2x+4\rightarrow3x+4 \\ (3x+4)-23 \\ 3x-19 \\ \text{equate them both sides and we have} \\ 2x-4=3x-19,\text{ left hand is the first part, right hand is the second part} \\ \text{solve for }x \\ 2x-4=3x-19 \\ -4+19=3x-2x \\ 15=x \\ x=15 \\ \text{so the three odd integers are, } \\ x=15 \\ x+2=17 \\ x+4=19 \\ 15,17,19 \end{gathered}[/tex]