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The vertices of a triangle ABC are A(3, 2, 0), B(5, 3, 2) and C(-9, 6, -3). The bisector AD of ∠ A meets BC at D, find the fourth vertex D.

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The given co-ordinates: A(3, 2, 0), B(5, 3, 2) and C(-9, 6, -3)

Now, AB = \(\sqrt{(5-3)^2+(3-2)^2+(2-0)^2}\) = \(\sqrt{4+1+4}\) = 3

Also, AC = \(\sqrt{(-9-3)^2+(6-2)^2+(-3-0)^2}\) = \(\sqrt{144+16+9}\)  = 13

Now, we have, AB/AC = 3/13

By the property of internal angle bisector,

AB/AC = BD/CD

Therefore, BD/CD = 3/13

Applying the section formula, we get,

D(x,y,z) = \((\frac{3\times5-9\times13}{3+13},\frac{3\times3+6\times13}{3+13},\frac{3\times2-3\times13}{3+13})\)

D(x,y,z) = \((-\frac{102}{16},\frac{87}{16},\frac{33}{16})\)

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