Think of a wrench tightening a nut on a bolt. Your arm puts a force clockwise on the end of the wrench and it rotates. This creates a torque on the nut, and since the nut is tightening, imagine that the torque vector points from the nut to the head of the bolt.
The equation for torque is t = r x F, where r is the length of the wrench and F is the force you're pulling on it with. To show that a = (bxc) = -(c x b), consider that the vectors F and r don't change direction or magnitude just because their order is switched in the equation. In other words, your arm is still tightening the nut, so the torque is still pointing in the same direction. So, you need a negative sign in front of F x r to make t point in the same direction as at first.
t = r x F = -(F x r)
Let P be the point at which the nut is fixed on this plane of the paper which moves perpendicular to the plane of paper on rotation obeying right hand system.
PA be the wrench on the plane of paper .
Let AB be the line on the plane of paper you apply a force F on the wrench making an angle PAB = x , which is measured in the sense of PAB (anti clock wise.)
Then the the moment of the force T about the point P
= PA X F= ( Perpedicular distance from the point P to the line AB)*|F|= PA*|F| sin(anglePAB) -> the bolt moves in a direction perpendicular to the plane of paper towards us. This similar to aXb =|a|*|b|sin (angle measured from a to b)
Now the bolt is made to rotate in BAP direction. The angle measured in this sense is oppsite of ABP. Hence angle BAP =-x.
b X a is similar to keeping the same magnitude of the force, the same lenth of PA but the sense direction of movement of the wrench is BAP . So the angle is BAP as opposed to the ABP. This makes nut move away in a direction perpendicular to the plane of paper.
Therefore bXa = |b||a| sin (angle measured from b to a)