I give three points:

[1] Yes; they are nonmagnetic  both in their elemental states as well in +2 oxidation states because:

In the elemental state, they possess[ ns^2 np^6] outermost configuration with no unpaired electron. Also after loosing two electrons, their  outermost configuration becomes   [ np^6](n=2,3,4,5--)  in +2 state . Either way, they do no possess any unpaired electron,i.e. with no resultant spin and thus they are diamagnetic.

[2] In +3 oxidation state( though very unstable; but that is another point to explain), their outermost confifuration will be[ np^5]. They  will possess one unpaired electron and thus are expected to be paramagnetic.

[3] Now as regards to their their momenta:

 S=1/2

L=1

J=L+S=1+1/2=3/2

 Pls note that for LESS THAN HALF FULL SUBSHELL ground state, J=L-S                 ( modulus). But  when the subshell is MORE THAN HALF FULL, the ground state becomes  J = L+S(modulus).  Since here the configuration np^5 is more than half full, apply J=L+S[ Russell Saunder Coupling]

Thank you sir.  You are exactly right.  If it is in 3+ state surely it will contribute magnetic  moment. My question is that, Is there any possibility to achieve 3+ state? If yes, what are all the factors which will  need to optimize the AE metals  3+ state (such as temperature, pressure,....etc)? Because we know that AE metals are unstable in 3+ state.

... also magnetic structure not only because of their unpaired electron, but the exchange interaction in their solid form (crystal structure).

Whether a material is ferromagnetic depends on the relative strengths of the exchange interaction with the other interactions, determine by their balance. Example; though Iron (α-Fe, BCC) is the best known as ferromagnetic, but the austenit (γ-Fe, FCC) is form of Iron in paramagnetic phase. So just changing the crystal structure can switch between ferromagnetism and paramagnetism.

In Iron (γ-Fe) the interaction is called direct exchange between Fe-Fe.

The partial substitution of x (Ca, Sr, Ba) in perovskite La(1-x)MnO3, could change the behaviour of the material from antiferromagnetic to ferrimagnetic, this is caused by the double exchange interaction between different Mn oxidation state in  Mn-O-Mn.

There are also Superexchange and RKKY-exchange.

Thank you Dr. Maykel Manawan.

I agree with you

But in this case of La(1-x)MnO3 (x =Ca, Sr, Ba) system, what is the role of x? X is ionic size dependent. If we substitute high ionic radii elements such as (x =Ca, Sr, Ba) to LMO system distortion will be more.  Because of the distortion,  exchange interaction is taking place between different Mn ions (Mn3+-Mn4+). So my opinion is that  oxidation state of (x =Ca, Sr, Ba) is not changing.

 In continuation of my answer at the first place:

 Energetically, it is not possible because of the OVERALL EFFECT  of  following three prosesses:

[1] Very large difference betweenI2[ 2nd  ionization Energy] and I3[3rd ionization Energy] 

I3 ( third  IE) values  of these metals are much higher than their corresponding I2( Secone IE values[ Be=1755  and 14810; Mg=1470,7730 kJ/mol –wikipedia].  So much difference in energies is not possible to attain at ordinary conditions of temperatures.

The reson for the vey large extra enegy ABSORBED  to remove the third electron to form M^3(g)[ M= Be, Mg, Ca---] from M^2 lies in fact the the M^2  AE ions have EXTRA STABLE COMPLETE OCTETS[ ns^2np^^ n=2,3--] and  thus the  removal of third electrons to form M^3(g) is   not possible energetically.

 [II] HYDRATION ENERGIES

   As is known, when we form  compounds in the aqueous medium, the enery is released  in the form of hydration enery.The amount of energy RELEASED to form  compounds of SMALL SIZED M^3(g) is more than that that released  to form compounds of  LARGE SIZED M^2(g). So this factor should favor the formation of M^3.

[III] MORE NUMBER OF BONDS

 If we to form M^3  compounds, there would form THREE BONDS while t M^2  will form TWO BONDS. As the  BOND FORMARTION IS A RELEASE IN ENEGY PHENOMENON, more energy will be released in forming M^3 compounds that those formed by M^2 compounds. So this factor should also favor formation of M^3 compound.

 Now first factor is ENERGY ABSORBING PROCESS while the other two processes are ENERGY RELEASING processes. BUT THE ENERGY ABSORBED IN THE FIRST PROCESS IS MUCH MORE THAN THE SUM OE ENERGIES RELEASED IN II AND III PROCESSES.

HENCE THE  CONVERSION OF M^2  To M^3 is  ENERGETICALLY UNFAVORABLE.

Dear Professor... Thank you very much for an interesting answer. Very kind of you Sir.

x is the partial substitution of Ca, Sr or Ba to La. Example LaMnO3 at room temperature, the super-exchange of Mn3+ - O - Mn3+ caused the chanted anti-ferromagnetic arrangement both Mn3+ (Exclusion Principle). While in La0.7Ca0.3MnO3, due to the substitution with Ca2+, the charge balance had to maintain with Mn4+, so the double-exchange of Mn3+ - O - Mn4+ have the ferrimagnetic arrangement. This effect is also dictate by the amount of substitution, above x=0.5 it become antiferromagnetic.

"So my opinion is that oxidation state of (x =Ca, Sr, Ba) is not changing." Yes U r rite.

Thank you very much Dr. Maykel Manawan for a wonderful discussion