3 Tangent spheres

Let $(x, y, z)$ be the center of sphere S3. We are given:

$-2x + y + 3z + 6 = 0 \quad (1)$

The distance between the centers of spheres S1 and S3 is $5 + 3 = 8$ , so:

$(x - 1)^2 + (y - 4)^2 + (z - 5)^2 = 8^2 \quad (2)$

and the distance between the centers of spheres S2 and S3 is $4 + 3 = 7$ , so:

$(x - 9)^2 + (y - 8)^2 + (z - 6)^2 = 7^2 \quad (3)$

Subtracting $(2) - (3)$ leads to $8x + 4y + z = 77$ , which combined with equation $(1)$ leads to $x = \frac{1}{26}(237-11y)$ and $z = \frac{1}{13}(53 - 8y)$ . Substituting these equations back into $(2)$ gives $(\frac{1}{26}(237-11y) - 1)^2 + (y - 4)^2 + (\frac{1}{13}(53 - 8y) - 5)^2 = 8^2$ , which solves to $y = \frac{1}{27}(119 \pm 2\sqrt{1351})$ .

The sum of the two $y$ -coordinates is therefore $\frac{2 \cdot 119}{27} \approx \boxed{8.815}$ .

Let the center of the sphere $S_3$ be $(h, k, l)$ . Then $h^2+k^2+l^2=2h+8k+10l+22=18h+16k+12l-132$ . From this we get $h=\dfrac{237-11k}{26}, l=\dfrac{53-8k}{13}$ . So $(\dfrac{237-11k}{26})^2+k^2+(\dfrac{53-8k}{13})^2=2\times \dfrac{237-11k}{26}+8k+10\times \dfrac{53-8k}{13}+22$ . Simplifying we get $1053k^2-9282k+12649=0$ . Hence the sum of the two $y$ coordinates is $\dfrac{9282}{1053}\approx \boxed {8.8148}$