Let [tex]x,y,z[/tex] denote the amounts (in liters) of the 20%, 30%, and 60% solutions used in the mixture, respectively.
The chemist wants to end up with 72 L of solution, so
[tex]x+y+z=72[/tex]
while using twice as much of the 60% solution as the 30% solution, so
[tex]z = 2y[/tex]
The mixture needs to have a concentration of 35%, so that it contains 0.35•75 = 26.25 L of pure acid. For each liter of acid solution with concentration [tex]c\%[/tex], there is a contribution of [tex]\frac c{100}[/tex] liters of pure acid. This means
[tex]0.20x + 0.30y + 0.60z = 26.25[/tex]
Substitute [tex]z=2y[/tex] into the total volume and acid volume equations.
[tex]\begin{cases}x+3y = 72 \\ 0.20x + 1.50y = 26.25\end{cases}[/tex]
Solve for [tex]x[/tex] and [tex]y[/tex]. Multiply both sides of the second equation by 5 to get
[tex]\begin{cases}x+3y = 72 \\ x + 7.50y = 131.25\end{cases}[/tex]
By elimination,
[tex](x+3y) - (x+7.50y) = 72 - 131.25 \implies -4.50y = -59.25 \implies \boxed{y=\dfrac{79}6} \approx 13.17[/tex]
so that
[tex]x+3\cdot\dfrac{79}6 = 72 \implies x = \boxed{\dfrac{65}2} = 32.5[/tex]
and
[tex]z=2\cdot\dfrac{79}6 = \boxed{\dfrac{79}3} \approx 26.33[/tex]
