(0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.

Let the points be

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

Firstly let us calculate the distance of PQ, OR and PR

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 =

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

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

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

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

Calculating PR

P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)

By using the formula,

Distance PR =

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

So here,

x_{1} = 0, y_{1} = 7, z_{1} = 10

x_{2} = – 4, y_{2} = 9, z_{2} = 6

Distance PR =

\( \sqrt{(-4 – 0)^2 + (9 – 7)^2 + (6 – 10)^2}\)

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

= \( \sqrt{16 + 4 + 16}\)

**= \( \sqrt{36}\)**

Now,

PQ^{2} + QR^{2} = 18 + 18

= 36

= PR^{2}

By using converse of Pythagoras theorem,

The given vertices P, Q & R are the vertices of a right – angled triangle at Q.

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