Let the mid points of the sides of a triangle ABC be ( 5 2, 7 ), ( 5 2, 3 ) and (4, 5). If its incentre is (h, k), then 3h + k is equal to :

Let the vertices of the triangle $ABC$ABC be $A(x_1, y_1)$A(x1​,y1​), $B(x_2, y_2)$B(x2​,y2​), and $C(x_3, y_3)$C(x3​,y3​).

Let the given midpoints be of sides $BC$BC, $CA$CA, and $AB$AB respectively:

Midpoint of $BC = \left(\dfrac{5}{2}, 7\right) \Rightarrow x_2 + x_3 = 5$BC=(25​,7)⇒x2​+x3​=5 and $y_2 + y_3 = 14$y2​+y3​=14

Midpoint of $CA = \left(\dfrac{5}{2}, 3\right) \Rightarrow x_3 + x_1 = 5$CA=(25​,3)⇒x3​+x1​=5 and $y_3 + y_1 = 6$y3​+y1​=6

Midpoint of $AB = (4, 5) \Rightarrow x_1 + x_2 = 8$AB=(4,5)⇒x1​+x2​=8 and $y_1 + y_2 = 10$y1​+y2​=10

Adding the three equations for the $x$x-coordinates:

$$2(x_1 + x_2 + x_3) = 5 + 5 + 8 = 18$$2(x1​+x2​+x3​)=5+5+8=18

$$x_1 + x_2 + x_3 = 9$$x1​+x2​+x3​=9

Substituting the known sums:

$$x_1 = 9 - 5 = 4$$x1​=9−5=4

$$x_2 = 9 - 5 = 4$$x2​=9−5=4

$$x_3 = 9 - 8 = 1$$x3​=9−8=1

Similarly, for $y$y-coordinates:

$$2(y_1 + y_2 + y_3) = 14 + 6 + 10 = 30$$2(y1​+y2​+y3​)=14+6+10=30

$$y_1 + y_2 + y_3 = 15$$y1​+y2​+y3​=15

$$y_1 = 15 - 14 = 1$$y1​=15−14=1

$$y_2 = 15 - 6 = 9$$y2​=15−6=9

$$y_3 = 15 - 10 = 5$$y3​=15−10=5

So, the vertices are $A(4, 1)$A(4,1), $B(4, 9)$B(4,9), and $C(1, 5)$C(1,5).

Lengths of the sides:

$$a = BC = \sqrt{(4 - 1)^2 + (9 - 5)^2} = \sqrt{9 + 16} = 5$$a=BC=(4−1)2+(9−5)2​=9+16​=5

$$b = CA = \sqrt{(4 - 1)^2 + (1 - 5)^2} = \sqrt{9 + 16} = 5$$b=CA=(4−1)2+(1−5)2​=9+16​=5

$$c = AB = \sqrt{(4 - 4)^2 + (9 - 1)^2} = \sqrt{0 + 64} = 8$$c=AB=(4−4)2+(9−1)2​=0+64​=8

Coordinates of the incentre $(h, k)$(h,k):

$$I(x, y) = \left(\dfrac{ax_1 + bx_2 + cx_3}{a + b + c}, \dfrac{ay_1 + by_2 + cy_3}{a + b + c}\right)$$I(x,y)=(a+b+cax1​+bx2​+cx3​​,a+b+cay1​+by2​+cy3​​)

$$h = \dfrac{5(4) + 5(4) + 8(1)}{5 + 5 + 8} = \dfrac{20 + 20 + 8}{18} = \dfrac{48}{18} = \dfrac{8}{3}$$h=5+5+85(4)+5(4)+8(1)​=1820+20+8​=1848​=38​

$$k = \dfrac{5(1) + 5(9) + 8(5)}{5 + 5 + 8} = \dfrac{5 + 45 + 40}{18} = \dfrac{90}{18} = 5$$k=5+5+85(1)+5(9)+8(5)​=185+45+40​=1890​=5

So, the incentre is $(h, k) = \left(\dfrac{8}{3}, 5\right)$(h,k)=(38​,5).

Therefore:

$$3h + k = 3 \cdot \dfrac{8}{3} + 5 = 8 + 5 = 13$$3h+k=3⋅38​+5=8+5=13

Hence, the correct option is $(3)\ 13$(3) 13.