<s>[INST] How to find the intersection of three spheres (full solutions)? The three equations of spheres are given
$(x-x_{1})^2+(y-y_{1})^2+(z-z_{1})^2=a^2$
$(x-x_{2})^2+(y-y_{2})^2+(z-z_{2})^2=b^2$
$(x-x_{3})^2+(y-y_{3})^2+(z-z_{3})^2=c^2$
How do I find $(x,y,z)$ analytically? 
 [/INST] To get you started:


*

*$(x-x_{1})^2+(y-y_{1})^2+(z-z_{1})^2=a^2$

*$(x-x_{2})^2+(y-y_{2})^2+(z-z_{2})^2=b^2$

*$(x-x_{3})^2+(y-y_{3})^2+(z-z_{3})^2=c^2$


Use 1.+2.


*

*$x^2-2xx_1+x_1^2+y^2-2yy_1+y_1^1+z_1^2-2zz_1+z_1^2=a^2$

*$x^2-2xx_2+x_2^2+y^2-2yy_2+y_2^2+z_2^2-2zz_2+z_2^2=b^2$


Now 1. - 2.
$$ -2xx_1+2xx_2+x_1^2-x_2^2-2yy_1+2yy_2+2zz_1-2zz_2+y_1^2-y_2^2+z_1^2-z_2^2=a^2-b^2$$
Simplify
$$2x(x_2-x_1)+2y(y_2-y_1)+2z(z_2-z_1)+(x_1^2-x_2^2+y_1^2-y_2^2+z_1^2-z_2^2)=a^2-b^2$$
$$\Leftrightarrow x=\dfrac{a^2-b^2-2y(y_2-y_1)-2z(z_2-z_1)-(x_1^2-x_2^2+y_1^2-y_2^2+z_1^2-z_2^2)}{2(x_2-x_1)}$$
Now put $x$ into 1. or 2. and solve the quadratic for $y$.
Use at last Ari's hint.
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