Let $$A(1, 0)$$, $$B(6, 2)$$ and $$C\left(\frac{3}{2}, 6\right)$$ be the vertices of a triangle ABC. If P is a point inside the triangle ABC such that the triangles APC, APB and BPC have equal areas, then the length of the line segment PQ, where Q is the point $$\left(-\frac{7}{6}, -\frac{1}{3}\right)$$, is

We are given the vertices of the triangle as $$A(1,0),\;B(6,2),\;C\left(\dfrac32,6\right).$$ Inside this triangle there is a point $$P$$ such that the areas of the three triangles $$\triangle APC,\;\triangle APB,\;\triangle BPC$$ are equal.

First recall a well-known fact: for any point $$P$$ inside a triangle, the ratios of the areas of the sub-triangles to the whole triangle give the barycentric coordinates of $$P$$. If all three areas are equal, each is one third of the total area, so the barycentric coordinates of $$P$$ are $$\left(\dfrac13,\dfrac13,\dfrac13\right).$$ A point with equal barycentric coordinates is nothing but the centroid of the triangle. Hence $$P$$ must be the centroid of $$\triangle ABC.$$

The centroid formula states that for vertices $$\bigl(x_1,y_1\bigr),\bigl(x_2,y_2\bigr),\bigl(x_3,y_3\bigr)$$ the centroid $$G(x_G,y_G)$$ is

$$x_G=\dfrac{x_1+x_2+x_3}{3},\qquad y_G=\dfrac{y_1+y_2+y_3}{3}.$$

Applying this to our triangle:

$$x_P=\dfrac{1+6+\dfrac32}{3}=\dfrac{1+6+1.5}{3}=\dfrac{8.5}{3}=\dfrac{17}{6},$$

$$y_P=\dfrac{0+2+6}{3}=\dfrac{8}{3}.$$

Thus $$P\left(\dfrac{17}{6},\dfrac83\right).$$

We are asked for the length of segment $$PQ$$ where $$Q\!\left(-\dfrac76,-\dfrac13\right).$$ We now use the distance formula between two points $$\bigl(x_1,y_1\bigr)$$ and $$\bigl(x_2,y_2\bigr):$$

$$\text{Distance}=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.$$

Compute the differences:

$$\Delta x = x_P - x_Q = \dfrac{17}{6}-\left(-\dfrac76\right)=\dfrac{17+7}{6}=\dfrac{24}{6}=4,$$

$$\Delta y = y_P - y_Q = \dfrac83-\left(-\dfrac13\right)=\dfrac{8+1}{3}=\dfrac{9}{3}=3.$$

Substituting these into the distance formula gives

$$PQ=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5.$$

So, the answer is $$5$$.