**Given:** Points are A(4, 0, 0) and B(-4, 0, 0)

**To find: **the locus of point P, the sum of whose distances from the given points is equal to 10, i.e. PA + PB = 10

Let the required point P(x, y, z)

**Formula used: **

The distance between any two points (a, b, c) and (m, n, o) is given by,

\(\sqrt{(a−m)^2+(b−n)^2+(c−o)^2}\)

The distance between P(x, y, z) and A(4, 0, 0) is PA,

=\(\sqrt{(x-4)^2+(y−0)^2+(z-0)^2}\)

=\(\sqrt{(x-4)^2+y^2+z^2}\)

Distance between P(x, y, z) and B(-4, 0, 0) is PB

=\(\sqrt{(x-(-4))^2+(y−0)^2+(z-0)^2}\)

= \(\sqrt{(x-3)^2+(y−2)^2+(z+1)^2}\)

=\(\sqrt{(x+4)^2+y^2+z^2}\)

According to question: PA + PB = 10

⇒ \(\sqrt{(x-4)^2+y^2+z^2}\) + \(\sqrt{(x+4)^2+y^2+z^2}\) = 10

⇒ \(\sqrt{(x-4)^2+y^2+z^2}\) =10- \(\sqrt{(x+4)^2+y^2+z^2}\)

Squaring both sides:

⇒ 16x^{2 }+ 625 – 100x = 25x^{2}+ 400 + 200x + 25y^{2 }+ 25z^{2}

⇒ 16x^{2}+ 625 – 100x – 25x^{2}– 400 – 200x – 25y^{2} – 25z^{2 }= 0

⇒ -9x^{2} – 25y^{2} – 25z^{2} – 300x + 225 = 0

⇒ 9x^{2} + 25y^{2} + 25z^{2} + 300x – 225 = 0

Hence locus of point P is 9x^{2} + 25y^{2} + 25z^{2} + 300x – 225 = 0