 Solution

For a real scalar triplet, with metric signature $(+---)$, differentiating the Lagrangian density gives $\partial \mathcal{L}/\partial {\varphi }_a=-\lambda {\varphi }_a$ and $\partial \mathcal{L}/\partial ({\partial }_{\mu }{\varphi }_a)={\partial }^{\mu }{\varphi }_a$. HenceFor a constant internal rotation parameter, the variation isEach bracketed tensor is symmetric in $a,c$, so its contraction with the antisymmetric symbol vanishes. This rotates the three field components, not the space coordinates; both the kinetic term and common mass term are invariant.

For a first-principles Noether current derivation, temporarily let $\theta$ depend on spacetime. The terms proportional to $\theta$ still cancel, leavingOn a solution of the field equations, the action variation for a compactly supported $\theta$ vanishes after integration by parts. Thus ${\partial }_{\mu }{j}^{\mu }=0$. To match the charges printed in the question, use the equivalent overall sign convention for the SO(3) Noether current of a real scalar tripletArbitrariness of $\mathbf{n}$ gives three continuity equations. Their charges are exactlyAn overall reversal of all currents is immaterial; the minus sign above keeps the generator convention distinct from the supplied charge convention.

Direct verification is also short:The mass term and the products of equal-type derivatives cancel by antisymmetry; integration by parts produces the boundary term. The charges are conserved when these integrals exist and the net boundary flux vanishes: sufficiently decaying fields on all space, periodic boundary conditions, or suitable symmetry-preserving no-flux boundaries suffice. Arbitrary open boundaries do not. A common constant $\lambda$ is necessary for this internal symmetry; its positivity is a stability issue, not needed for the algebraic conservation proof.