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#