Ans: b) $\frac{1}{6}$AB.

Soln:

Let PB=r=PQ. QR=x. In triangle PRB, RB=2r-x. PR= r+x.

Since PRB is right angled, $(r+x{)}^2=(2r-x{)}^2+r^2$ .

Solving we get x=$\frac{r}{3}$=$\frac{AB}{6}$

option (b)

a is false as if one side is 1, another side is a, then the third side < a+1. this is not possible as it needs to be an integer value and since it’s a scalene triangle, the third side cannot be 1+a

b is true

c is false as there is a possibility of 120${}^{\circ }$, 30${}^{\circ }$ , 30${}^{\circ }$

Answer- (d) None of these

Solution:- Calculate the altitude of the equilateral triangle. Then according to the conditions given,
Diameter of the circle circumscribing the equilateral triangle= Altitude of the equilateral triangle + Diameter of the smaller cirlce.
Solve the above equation for the required answer. 

Alternate Soln: The triangle’s dimension is used to calculate the radii of the outermost and the second circle.

Radius of the bigger circle: R=$\frac{\left(1.5\right)}{cos30}$; =$\sqrt{3}$ units.

Radius of smaller circle:

r = 1.5 tan 30. =  $\frac{\frac{3}{2}}{\sqrt{3}}$ r= $\frac{\sqrt{3}}{2}$.

Radius of smallest circle:

$\frac{\left(\right(radiusofbiggest)-(radiusof{2}^{nd}\left)\right)}{2}$;

= $\frac{\sqrt{3}}{4}$

Hence ans: d (none of these)