If $$3x^2 - ax + 9 = ax^2 + 2x + 5$$ has only one (repeated) solution, then the positive integral solution of $$a$$ is

$$3(x^2)-ax+9 = a(x^2)+2x+5$$

$$(x^2)(3-a)-x(a+2)+4=0$$

If a quadratic equation$$(ax^2+bx+c=0) $$ has equal roots, then discriminant should be zero i.e.$$ b^2-4ac=0$$

Here $$ a=3-a ; b=-(a+2) ; c=4 $$

$$(-(a+2))^2 - 4(3-a)\times4 = 0$$

$$a^2 + 4 + 4a - 16(3-a)=0$$

$$a^2 + 4 + 4a - 48 + 16a = 0$$

$$a^2 + 20a - 44 = 0$$

$$(a+22)(a-2)=0$$

$$a=2$$

When a=2, equation becomes $$(x^2)(3-2)-x(2+2)+4=0$$

$$x^2-4x+4 =0$$

$$x=2$$

Option D is correct.