Line perpendicular to y=2/3+3 and passes through 2,1

I'm assuming there is an x after 2/3?

y = (2/3)x + 3

First we invert the gradient to make it perpendicular.

y = (-3/2)x + 3

Now we need to make y equal to 1, when x is 2. Currently, y is 0, since (-3/2) * 2 is -3, plus 3 is 0. We could just add 4 instead of 3 to make y equal to 1, so the line...

y = (-3/2)x + 4

is perpendicular to the line you provided, and passes through (2, 1).