Okay, you solve the two equation for x, in terms of y and z, and, setting those equal, get \(\displaystyle 2+ \frac{5}{3}y- \frac{4}{3}z= \frac{4}{3}+ y- \frac{2}{3}z\). That's correct. You then solve for z, in terms of y, first by subtracting \(\displaystyle \frac{5}{3}y\) from both sides. That's where you make your mistake. On the right, you have y and subtract \(\displaystyle \frac{5}{3}y\) from that. That is \(\displaystyle \left(1- \frac{5}{3}\right)y= \left(\frac{3}{3}- \frac{5}{3}\right)y= -\frac{2}{3}y\) but in your next line you have just \(\displaystyle -\frac{5}{3}y\). Then subtracting 2 from both sides, you should have \(\displaystyle -\frac{2}{3}z= \frac{2}{3}- \frac{2}{3}y\) (again that is 2/3 times y, not 5/3).
Finally, dividing both sides by \(\displaystyle -\frac{2}{3}\), \(\displaystyle z= -1+ y\).
Putting that into the original equation of the first plane, \(\displaystyle 3x- 5y+ 4z= 3x- 5y+ 4(y- 1)= 3x- y- 4= 6\) or \(\displaystyle y= 3x+ 2\). Since I don't like fractions I will use x rather than y as parameter: x= t, y= 3t+ 2, z= y-1= 3t+ 2-1= 3t+ 1.