The normal has an equation
6y + 2x = -1 (given).
Its slope is -1/3.
It means that the tangent line to the parabola has the slope m= 3. From the given equation of the parabola, the slope of the tangent line at abscissa x is (1/2) * 3 * 2x = 3x (after taking the derivative).
In order for the slope of a tangent line 3x be m= 3, x should be equal to 1.
Thus we found out that x-coordinate of point A is x= 1.
Substitute x= 1 into equation of the normal line 6y + 2x = -1 . You will get 6y + 2*1 = -1, or 6y = -1 - 2 = -3, which gives y = -3/6 = -1/2.
Thus we learned that tangent point A on the parabola has coordinates A = (x,y) = (1, -1/2).
Now we know that the parabola passes throw the point (1, -1/2).
From equation of the parabola, it means that
-1/2 = (1/2) * (3 * 12 - k).
It gives
-1 = 3 - k,
which implies
k = 3 + 1 = 4.
k = 4.
Part ( i ) is completed in full.
Part ( ii ) is simple arithmetic.