Let us define a function

$$f\left( x \right) = {{ax{}^3} \over 3} + {{b{x^2}} \over 2} + cx$$

Being polynomial, it is continuous and differentiable, also,

$$f\left( 0 \right) = 0\,$$ and $$\,\,f\left( 1 \right) = {a \over 3} + {b \over 2} + c$$

$$ \Rightarrow f\left( 1 \right) = {{2a + 3b + 6c} \over 6} = 0$$ (given)

$$\therefore$$ $$f\left( 0 \right) = f\left( 1 \right)$$

$$\therefore$$ $$f(x)$$ satisfies all conditions of Rolle's theorem

therefore $$f'\left( x \right) = 0$$ has a root in $$\left( {0,1} \right)$$

i.e. $$a{x^2} + bx + c = 0$$ has at lease one root in $$(0, 1)$$

$$f''\left( x \right) = 6\left( {x - 1} \right).$$ Inegrating,

we get $$f'\left( x \right) = 3{x^2} - 6x + c$$

Slope at $$\left( {2,1} \right) = f'\left( 2 \right) = c = 3$$

$$\left[ {\,\,} \right.$$ As slope of tangent at $$(2, 1)$$ is $$3$$ $$\left. {\,\,} \right]$$

$$\therefore$$ $$f'\left( x \right) = 3{x^2} - 6x + 3 = 3{\left( {x - 1} \right)^2}$$

Inegrating again, we get

$$f\left( x \right) = {\left( {x - 1} \right)^3} + D$$

The curve passes through $$(2,1)$$

$$ \Rightarrow 1 = {\left( {2 - 1} \right)^3} + D \Rightarrow D = 0$$

$$\therefore$$ $$f\left( x \right) = {\left( {x - 1} \right)^3}$$

$${y^2} = 18x \Rightarrow 2y{{dy} \over {dx}} = 18 \Rightarrow {{dy} \over {dx}} = {9 \over y}$$

Given $${{dy} \over {dx}} = 2 \Rightarrow {9 \over 2} = 2 \Rightarrow y = {9 \over 2}$$

Puting in $${y^2} = 18x \Rightarrow x = {9 \over 8}$$

$$\therefore$$ Required point is $$\left( {{9 \over 8},{9 \over 2}} \right)$$

$$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1$$

$$f'\left( x \right) = 6{x^2} - 18ax + 12{a^2};$$

$$\,\,\,\,\,\,\,\,\,\,f''\left( x \right) = 12x - 18a$$

For max. or min.

$$6{x^2} - 18ax + 12{a^2} = 0$$

$$ \Rightarrow {x^2} - 3ax + 2{a^2} = 0$$

$$ \Rightarrow x = a$$ or $$x=2a.$$

At $$x=a$$

$$f''(a) = 12a - 18a = -6a < 0$$

At $$x=a$$ maximum As f''($$a$$) < 0.

At $$x=2a$$

$$f''(a) = 24a - 18a = 6a > 0$$

At $$x=2a$$ minimum As f''(2$$a$$) > 0.

$$\therefore$$ $$p=a$$ and $$q=2a$$

As per question $${p^2} = q$$

$$\therefore$$ $${a^2} = 2a $$

$$ \Rightarrow $$ $$a(a - 2) = 0$$

$$ \therefore $$ $$a$$ = 0, 2

but $$a > 0,$$ therefore, $$a=2.$$