(0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.

Let us consider the points be

P(0, 7, –10), Q(1, 6, – 6) and R(4, 9, – 6)

If any 2 sides are equal, hence it will be an isosceles triangle

So firstly let us calculate the distance of PQ, QR

Calculating PQ

P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)

By using the formula,

Distance PQ =

\( \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)

So here,

x_{1} = 0, y_{1} = 7, z_{1} = – 10

x_{2} = 1, y_{2} = 6, z_{2} = – 6

Distance PQ =

\( \sqrt{(1 – 0)^2 + (6 – 7)^2 + (-6 – (-10))^2}\)

= \( \sqrt{(1)^2 + (-1)^2 + (4)^2}\)

= \( \sqrt{1 + 1 + 16}\)

**= \( \sqrt{18}\)**

Calculating QR

Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)

By using the formula,

Distance QR =

\( \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)

So here,

x_{1} = 1, y_{1} = 6, z_{1} = – 6

x_{2} = 4, y_{2} = 9, z_{2} = – 6

Distance QR =

\( \sqrt{(4 – 1)^2 + (9 – 6)^2 + (-6 – (-6))^2}\)

= \( \sqrt{(3)^2 + (3)^2 + (-6+6)^2}\)

= \( \sqrt{9 + 9 + 0}\)

**= \( \sqrt{18}\)**

Hence, PQ = QR

18 = 18

2 sides are equal

PQR is an isosceles triangle.

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