Given that the zeroes of the cubic polynomial $x^{3}-6x^{2}+3x+10$ are of the form $a, a+b, a+2b$ for some real numbers $a$ and $b$,find the values of $a$ and $b$ as well as the zeroes of the given polynomial.

(A) Let $f(x) = x^{3}-6x^{2}+3x+10$.
Given that $a, (a+b),$ and $(a+2b)$ are the zeroes of $f(x)$.
Sum of the zeroes $= -\frac{\text{Coefficient of } x^{2}}{\text{Coefficient of } x^{3}} = -\frac{-6}{1} = 6$.
So,$a + (a+b) + (a+2b) = 6 \Rightarrow 3a + 3b = 6 \Rightarrow a+b = 2 \Rightarrow b = 2-a \dots (i)$.
Product of the zeroes $= -\frac{\text{Constant term}}{\text{Coefficient of } x^{3}} = -\frac{10}{1} = -10$.
So,$a(a+b)(a+2b) = -10$.
Substituting $a+b=2$ and $b=2-a$,we get $a(2)(a+2(2-a)) = -10$.
$2a(a+4-2a) = -10 \Rightarrow 2a(4-a) = -10 \Rightarrow 8a - 2a^{2} = -10$.
$2a^{2} - 8a - 10 = 0 \Rightarrow a^{2} - 4a - 5 = 0$.
$(a-5)(a+1) = 0$.
Thus,$a = 5$ or $a = -1$.
If $a = 5$,then $b = 2-5 = -3$. The zeroes are $5, 5-3, 5-6$,i.e.,$5, 2, -1$.
If $a = -1$,then $b = 2-(-1) = 3$. The zeroes are $-1, -1+3, -1+6$,i.e.,$-1, 2, 5$.
Thus,the values are $(a=5, b=-3)$ or $(a=-1, b=3)$,and the zeroes are $-1, 2, 5$.