Solve for x: 3|x-3| 2=14
a. no solution
b. x=-1,x=8.3
c. x=0,x=7
d. x=-1,x=7

I am looking  on the answers, and there is only one case, when a or b or c or d pass:   3|x-3| + 2 = 14.  So I assume, that before two is plus. Then:

3|x-3|+2=14    |minus 2
3|x-3|=12        |divide 3
|x-3|=4 

From absolute value definition you've got two ways:

x-3=4     or     x-3=-4
x=7        or     x=-1

And  answer d) passes