Substituting #x=3# in the function #f(x)=3(x-2)^2-5x+2#, we get y-coordinate of point as follows
#y=f(3)#
#=3(3-2)^2-5\cdot 3+2#
#=-10#
The slope #dy/dx# of tangent to the curve: #f(x)=3(x-2)^2-5x+2# is given by differentiating the function #f(x)# w.r.t. #x# as follows
#f'(x)=d/dx(3(x-2)^2-5x+2)#
#=6(x-2)-5#
Now, the slope of tangent at #x=3#,
#f'(3)=6(3-2)-5#
#=1#
hence, the slope #m# of normal at #(3, -10)# is given as
#m=-1/{f'(3)}=-1/1=-1#
Now, equation of the normal at the point #(x_1, y_1)\equiv(3, -10)# & having sloe #m=-1# is given by following formula
#y-y_1=m(x-x_1)#
#y-(-10)=-1(x-3)#
#x+y+7=0#