$\left(i\right)$ Verify the points $(0,7,-10),(1,6,-6)\text{and}(4,9,-6)$ are the vertices of an isosceles triangle.

$\left(ii\right)$ Verify the points $(0,7,10),(-1,6,6)\text{and}(-4,9,6)$ are the vertices of a right angled triangle.

$\left(iii\right)$ Verify the points $(-1,2,1),(1,-2,5),(4,-7,8)\text{and}(2,-3,4)$ are the vertices of a parallelogram.

$\left(i\right)$ Finding side lengths.

Let the given points be
$A(0,7,-10),B(1,6,-6)\&C(4,9,-6)$
Now, we calculate the side lengths $AB,BC\&AC$

Calculating $AB$:

$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

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

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

Calculating $BC$:

$BC=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

$=\sqrt{(4-1{)}^2+(9-6{)}^2+(-6-(-6){)}^2}$
$=\sqrt{(3{)}^2+(3{)}^2+(-6+6{)}^2}$

$=\sqrt{9+9+0}$
$=\sqrt{18}$
$=3\sqrt{2}$

$AB=3\sqrt{2},BC=3\sqrt{2}$
Verification
$\because AB=BC$
Hence, $ABC$ is an isosceles triangle.

Note: We don't need to calculate third side, because for an isosceles triangle we need only two sides to be equal.

$\left(ii\right)$ Finding side lengths.

Let the given points be
$A(0,7,10),B(-1,6,6)\&C(-4,9,6)$
Now, we calculate the side lengths $AB,BC\&AC$

Calculating $AB$:

$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

$=\sqrt{(-1-0{)}^2+(6-7{)}^2+(6-10{)}^2}$
$=\sqrt{(-1{)}^2+(-1{)}^2+(-4{)}^2}$

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

Calculating $BC$:
$B(-1,6,6)\&C(-4,9,6)$

$BC=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

$=\sqrt{(-4-(-1){)}^2+(9-6{)}^2+(6-6{)}^2}$
$=\sqrt{(-4+1{)}^2+(3{)}^2+(0{)}^2}$

$=\sqrt{9+9}$
$=\sqrt{18}$
$AB=\sqrt{18},BC=\sqrt{18}$

Calculating $AC$:
$A(0,7,10)\&C(-4,9,6)$

$AC=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

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

$=\sqrt{16+4+16}$
$=\sqrt{36}$
$=6$
$AB=\sqrt{18},BC=\sqrt{18},AC=6$

Verification
$A{B}^2+B{C}^2=(\sqrt{18}{)}^2+(\sqrt{18}{)}^2$
$=36$
$=A{C}^2$

As we know in Right angled triangle,

$(\text{Hypotenuse}{)}^2=(\text{Height}{)}^2+(\text{Base}{)}^2$

Here, $A{B}^2+B{C}^2=A{C}^2$

Hence, the given points form a right angled triangle.

$\left(iii\right)$ Finding lengths of sides and diagonal.
Let the given points be
$A(-1,2,1),B(1,-2,5)C(4,-7,8)\&D(2,-3,4)$

Calculating $AB$:

$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

$=\sqrt{(1-(-1){)}^2+(-2-2{)}^2+(5-1){)}^2}$
$=\sqrt{(2{)}^2+(-4{)}^2+(4{)}^2}$

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

Calculating $BC$:
$B(1,-2,5)\&C(4,-7,8)$

$BC=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

$=\sqrt{(3{)}^2+(-5{)}^2+(3{)}^2}$
$=\sqrt{9+25+9}$

$=\sqrt{43}$

$AB=6,BC=\sqrt{43}$

Calculating $CD$:
$C(4,-7,8)\&D(2,-3,4)$

$CD=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

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

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

$AB=6,BC=\sqrt{43},CD=6$

Calculating $AD$
$A(-1,2,1)\&D(2,-3,4)$
$=\sqrt{(2-(-1){)}^2+(-3-2{)}^2+(4-1{)}^2}$
$=\sqrt{(3{)}^2+(-5{)}^2+(3{)}^2}$

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

Now, $AB=CD$ and
$BC=AD$

Hence, Opposite sides are equal.

$AB=6,BC=\sqrt{43},CD=6,AD=\sqrt{43}$

Now, finding length of diagonals

Calculating $AC$
$A(-1,2,1)\&C(4,-7,8)$
$=\sqrt{(4-(-1){)}^2+(-7-2{)}^2+(8-1{)}^2}$
$=\sqrt{(5{)}^2+(-9{)}^2+(7{)}^2}$

$=\sqrt{25+81+49}$
$=\sqrt{155}$
$AC=\sqrt{155}$

Calculating $BD$
$B(1,-2,5)\&D(2,-3,4)$
$=\sqrt{(2-1){)}^2+(-3-(-2){)}^2+(4-5{)}^2}$
$=\sqrt{(1{)}^2+(-1{)}^2+(-1{)}^2}$

$=\sqrt{1+1+1}$
$=\sqrt{3}$

So, $AC\ne BD$ i.e., diagonal are not equal.

Final Answer:

The diagonals are not equal, but opposite sides are equal.

So, $ABCD$ is parallelogram.