All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.
(i)
\frac{dy}{dx} = 3x^{2} - 2ax + 3

When x = 1, ~\frac{dy}{dx} = 6 - 2a

m_{normal} = -\frac{1}{6-2a}

(ii)
Equation of normal: y - (10-a) = - \frac{1}{6-2a} (x-1)

Subs (-5, 3),

\Rightarrow a - 7 = \frac{6}{6-2a}

a^{2} - 10a + 24 = 0

(a-4)(a-6) = 0

a = 4 \mathrm{~or~} 6.

(iii)
If a =4, Normal: y - 6 = 0.5 (x-1)

\therefore, x = 11

Required coordinates = (11,11)

KS Comments

Some students left out the minus sign when converting to gradient of normal. Also, the answers need to be left in coordinates form.

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