Let the function f(x) = $$(x^{2}+1) |x^{2}-ax+2|+\cos|x|$$ be not differentiable at the two points x = $$\alpha$$ = 2 and $$x= \beta$$. Then the distance of the point $$(\alpha , \beta)$$ from the line $$12x+5y+10=0$$ is equal to :

The function is $$f(x) = (x^2+1)|x^2-ax+2| + \cos|x|$$. Note that $$\cos|x| = \cos x$$, which is differentiable everywhere, so any non-differentiability arises solely from the factor $$|x^2 - ax + 2|$$ and occurs where $$x^2 - ax + 2 = 0$$.

Since $$x = \alpha = 2$$ is given as a point of non-differentiability, it must satisfy $$4 - 2a + 2 = 0$$, which yields $$a = 3$$. Substituting this into the quadratic gives $$x^2 - 3x + 2 = (x-1)(x-2) = 0$$, so the roots are $$x = 1$$ and $$x = 2$$.

At $$x = 2$$, we have $$(x^2+1) = 5 \ne 0$$, confirming that the absolute value causes non-differentiability and hence $$\alpha = 2$$. At $$x = 1$$, we have $$(x^2+1) = 2 \ne 0$$, which similarly yields the other point of non-differentiability $$\beta = 1$$. Thus $$(\alpha,\beta) = (2,1)$$.

To find the distance from $$(2,1)$$ to the line $$12x + 5y + 10 = 0$$, we use the formula

$$ d = \frac{|12(2) + 5(1) + 10|}{\sqrt{144 + 25}} = \frac{|24 + 5 + 10|}{13} = \frac{39}{13} = 3 $$.

Therefore, the correct answer is Option 3: 3.