Yes, and I see what you mean. The two p orbitals when added together must have a nodal plane. So there must be a plane in which the electron density is zero.
Suppose we take the x axis to be along the triple bond, then look in the yz plane through one of the nuclei i.e. we're looking at a plane normal to the bond.
The electron density is proportional to $|\psi|^2$. Yes?
When we add $p_y$ and $p_z$ we get a function that still has the nodal plane.
That's the middle diagram.
The right diagram shows the result when we square $p_y + p_z$. Now this is positive everywhere because anything squared is positive everywhere, and there is still a nodal plane. OK so far?
In the middle diagram $p_y+p_z$ is antisymmetric about the nodal plane i.e. if we reflect in the nodal plane the magnitude is the same but we get a sign change.
@JohnRennie I'll consider this to be true for the time being sir. Even though I didn't understand this, I could understand your immediately above statement. So according to this, the orbital (region of maximum probability of finding an electron) looks much different from that of the third grey lobes diagram. Am I right, sir?
Ok sir. So it's meaningless to call an electron labelled "A" can be present in either positive lobe or negative lobe at different points of time.
I understand that we couldn't fix a momentum or position of an electron accurately as per Heisenberg's uncertainty principle. Is this why it's meaning less, sir?
