Verify the following:

Solution:

(i) Let points (0, 7, –10), (1, 6, –6), and (4, 9, –6) be denoted by A, B, and C respectively.

$\mathrm{AB}=\sqrt{(1-0)^{2}+(6-7)^{2}+(-6+10)^{2}}$

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

$=\sqrt{1+1+16}$

$=\sqrt{18}$

$=3 \sqrt{2}$

$\mathrm{BC}=\sqrt{(4-1)^{2}+(9-6)^{2}+(-6+6)^{2}}$

$=\sqrt{(3)^{2}+(3)^{2}}$

$=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$

$\mathrm{CA}=\sqrt{(0-4)^{2}+(7-9)^{2}+(-10+6)^{2}}$

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

$=\sqrt{16+4+16}=\sqrt{36}=6$

Here, AB = BC ≠ CA

Thus, the given points are the vertices of an isosceles triangle.

(ii) Let (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) be denoted by A, B, and C respectively.

$\mathrm{AB}=\sqrt{(-1-0)^{2}+(6-7)^{2}+(6-10)^{2}}$

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

$=\sqrt{1+1+16}=\sqrt{18}$

$=3 \sqrt{2}$

$\mathrm{BC}=\sqrt{(-4+1)^{2}+(9-6)^{2}+(6-6)^{2}}$

$=\sqrt{(-3)^{2}+(3)^{2}+(0)^{2}}$

$=\sqrt{9+9}=\sqrt{18}$

$=3 \sqrt{2}$

$\mathrm{CA}=\sqrt{(0+4)^{2}+(7-9)^{2}+(10-6)^{2}}$

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

$=\sqrt{36}$

$=6$

Now, $\mathrm{AB}^{2}+\mathrm{BC}^{2}=(3 \sqrt{2})^{2}+(3 \sqrt{2})^{2}=18+18=36=\mathrm{AC}^{2}$

Therefore, by Pythagoras theorem, ABC is a right triangle.

Hence, the given points are the vertices of a right-angled triangle.

(iii) Let (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) be denoted by A, B, C, and D respectively.

$\mathrm{AB}=\sqrt{(1+1)^{2}+(-2-2)^{2}+(5-1)^{2}}$

$=\sqrt{4+16+16}$

$=\sqrt{36}$

$=6$

$B C=\sqrt{(4-1)^{2}+(-7+2)^{2}+(8-5)^{2}}$

$=\sqrt{9+25+9}=\sqrt{43}$

$\mathrm{CD}=\sqrt{(2-4)^{2}+(-3+7)^{2}+(4-8)^{2}}$

$=\sqrt{4+16+16}$

$=\sqrt{36}$

$=6$

$\mathrm{DA}=\sqrt{(-1-2)^{2}+(2+3)^{2}+(1-4)^{2}}$

$=\sqrt{9+25+9}=\sqrt{43}$

Here, $A B=C D=6, B C=A D=\sqrt{43}$

Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.

Therefore, ABCD is a parallelogram.

Hence, the given points are the vertices of a parallelogram.