The side BC of a right-angled triangle ABC ($$\angle ABC = 90^\circ$$) is divided into four equal parts at P, Q and R respectively. If $$AP^2 + AQ^2 + AR^2 = 3 b^2 + 17 na^2$$ then n is equal to:

In$$ \triangle ABC, AB=c, BC=a, and AC= b$$

and BP=PQ=QR=RC=$$\frac{a}{4}$$

Now, by pythagoras theorem

$$(AP)^{2}=(AB)^{2}+(BP)^{2}=c^{2}+\left(\frac{a}{4}\right)^{2}$$ ---(i)

$$(AQ)^{2}=(AB)^{2}+(BQ)^{2}=c^{2}+\left(\frac{2a}{4}\right)^{2}$$---(ii)

$$(AR)^{2}=(AB)^{2}+(BR)^{2}=c^{2}+\left(\frac{3a}{4}\right)^{2}$$ ----(iii)

Adding (i), (ii) and (iii)

$$(AP)^{2}+(AQ)^{2}+(AR)^{2}=c^{2}+\frac{a^{2}}{16}+\frac{4a^{2}}{16}+\frac{9a^{2}}{16}$$

$$(AP)^{2}+(AQ)^{2}+(AR)^{2}=3c^{2}+\frac{14a^{2}}{16}=3c^{2}+\frac{7a^{2}}{8}$$

From, $$ \triangle ABC$$,

$$b^{2}=c^{2}+a^{2}$$

$$c^{2}=b^{2}-a^{2}$$

$$(AP)^{2}+(AQ)^{2}+(AR)^{2}=3c^{2}+\frac{7a^{2}}{8}$$

$$(AP)^{2}+(AQ)^{2}+(AR)^{2}=3(b^{2}-a^{2})+\frac{7a^{2}}{8}=3b^{2}-\left(3-\frac{7}{8}\right)a^{2}$$

$$(AP)^{2}+(AQ)^{2}+(AR)^{2}=3b^{2}-\frac{17}{8}a^{2}$$

Now, we will compare,

$$3b^{2}-\frac{17}{8}a^{2}=3 b^2 + 17 na^2$$

Hence, $$n = \frac{1}{8}$$