Given that ;
AS = 10cm
SN = 5cm
TN = 8cm
In the triangle ABC with the sides abc : the sum of the length of any two sides of the triangle is greater than the length of the third considering ANS $$5+AN=10$$
$$AN<15$$
$$AN>5$$
$$5<AN<15$$
$$Hence,\ \ AN=6,7,\ ...............,\ 14$$
Minimum value of AN=6 Maximum value =14
Using the minimum and maximum value of AN in triangle ATN
$$AT>TN-AN\left(\min imum\ value\ of\ AT\right)$$
$$AT>8-6\left(where\ 6\ =\min\ no\ of\ AN\right)$$
$$AT>2$$
$$AT>2\left(\min\ value\ of\ AT\right)$$
$$AT<AN+TN\left(Maximum\ value\ of\ AT\right)$$
$$AT<14+8\left(where\ 8\ is\ the\ \max imum\ value\ of\ AN\right)$$
$$AT<22$$
$$Maximum\ value\ of\ AT\ is\ 21$$
$$Positive\ difference\ =21-3=18$$
$$answer\ is\ Option\ B$$
