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We consider two spheres. The sphere of radius 3 centered at the origin (0, 0, 0). The second sphere is tangent to the first sphere and is centered at A(3, 5, 2). Find the point of intersection P of these spheres.

Hint: The points A, P and the origin are on the same line.

This means that the line of the forms 3Ax + 5Ay + 2Az contains both the centers.

This can be seen since At A=0, it contains (0,0,0) and at A=1, it contains (3,5,2).

The length along the line from A = 0 out to A = t is

is sqrt((3t)² + (5t)² + (2t)²).

Since 3 squared, 5 squared, and 2 squared are, respectively, 9, 25, and 4,

and 9+25+4 = 38, this is the same as sqrt(38t²).

The edge of the inside sphere is then at where the equation 38t² = 3² is satisfied.

That is, t² = 9/38, or t = 3*sqrt(38)/38. The point is (9,15,6)*sqrt(38)/38.

That gives the value (1.4600, 2.4333, 0.9733).

Since this is almost half the distance to the outside sphere,

the radius of the outside sphere would be a little bit bigger than the inside.

It would be roughly (1.5400² + 2.5667² + 1.0267²), which is roughly 3.16.

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