Distance in 3D Space

Consider points (1, 2, 3) and (4, 5, 6) exist in a three-dimensional space. What is the distance between the two points?

Hint
In a three-dimensional space, the distance between two points is
$$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$$
Hint 2
Since the difference between coordinates is squared, it doesn’t matter if Point 1 is assigned $$(x_1, y_1, z_1)$$ or $$(x_2, y_2, z_2)$$ as long as Point 2 is assigned as the other set.
In a three-dimensional space, the distance between two points is
$$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$$
Since the difference between coordinates is squared, it doesn’t matter if Point 1 is assigned $$(x_1, y_1, z_1)$$ or $$(x_2, y_2, z_2)$$ as long as Point 2 is assigned as the other set. Let’s arbitrarily set $$(1, 2, 3)$$ as $$(x_1, y_1, z_1)$$ :
$$$d=\sqrt{(4-1)^2+(5-2)^2+(6-3)^2}$$$
$$$=\sqrt{(3)^2+(3)^2+(3)^2}=\sqrt{9+9+9}=\sqrt{27}=5.2$$$
5.2