Q.Show that the points $A(1,2,3)$, $B(3,4,5)$ and $C(-1,0,1)$ are collinear.

Every point of space is located by an ordered triple $(x,y,z)$, its signed perpendicular distances from the three coordinate planes. Applying the Pythagorean theorem twice -- first across a coordinate plane, then out of it -- gives the distance between two points $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ as $P_1P_2=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$, the natural 3D extension of the 2D distance formula. This single formula also tests collinearity of three points, since three collinear points always satisfy the additive relation that the sum of the two smaller pairwise distances equals the largest.