A challenging problem of regular heptagon by HC Rajpoot

The solid angle subtended by the yellow triangle $OA_{-1}A_0$ at the point with coordinates $(u,v,25)$ is $\Omega(u,v) \; = \; \int_0^{6\cot\frac{\pi}{7}}\,dx \int_{-x\tan\frac{\pi}{7}}^{x\tan\frac{\pi}{7}} \frac{25}{\big(625 + (u-x)^2 + (v-y)^2\big)^{\frac32}}\,dy$ Now the solid angle subtended by the complete heptagon at the point with coordinates $(-6,0,25)$ (a point a height of $25$ above $A_3$ ) is equal to the sum of the solid angles subtended by the yellow triangle at the points $B_j$ with coordinates $(6\mathrm{cosec}\tfrac{\pi}{7}\cos\tfrac{(2n+1)\pi}{7},6\mathrm{cosec}\tfrac{\pi}{7}\sin\tfrac{(2n+1)\pi}{7},25)$ for $-3 \le j \le 3$ (the points which are a height of $25$ above the points $A_j$ for $-3 \le j \le 3$ . Thus the desired solid angle is $\sum_{j=-3}^3 \Omega\big(6\mathrm{cosec}\tfrac{\pi}{7}\cos\tfrac{(2n+1)\pi}{7},6\mathrm{cosec}\tfrac{\pi}{7}\sin\tfrac{(2n+1)\pi}{7}\big) \; = \; \boxed{0.523322685}$ according to Mathematica using WorkingPrecision set to $20$ .

A search of the Internet found this formula from the author's book. I used with $a=12$ and $h=25$ , and obtained a different answer... A typo, somewhere, perhaps.