= Solution
For a <real scalar triplet>, with metric signature $(+---)$, differentiating the <Lagrangian density> gives $\partial\mathcal L/\partial\phi_a=-\lambda\phi_a$ and $\partial\mathcal L/\partial(\partial_\mu\phi_a)=\partial^\mu\phi_a$. Hence
$$
\boxed{(\Box+\lambda)\phi_a=0,\qquad
\partial_t^2\phi_a-\nabla^2\phi_a+\lambda\phi_a=0.}
$$
For a constant internal <rotation> parameter, the variation is
$$
\delta\mathcal L=\theta\epsilon_{abc}n_b
\left[(\partial_\mu\phi_a)(\partial^\mu\phi_c)-\lambda\phi_a\phi_c\right]=0.
$$
Each 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, leaving
$$
\delta\mathcal L=j^\mu(\mathbf n)\partial_\mu\theta,\qquad
j^\mu(\mathbf n)=\partial^\mu\phi_a\epsilon_{abc}n_b\phi_c.
$$
On 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 triplet>
$$
\boxed{J_a^\mu=\epsilon_{abc}(\partial^\mu\phi_b)\phi_c,\qquad
j^\mu(\mathbf n)=-n_aJ_a^\mu.}
$$
Arbitrariness of $\mathbf n$ gives three <continuity equations>. Their charges are exactly
$$
\boxed{Q_a=\int d^3x\,J_a^0=\int d^3x\,\epsilon_{abc}\dot\phi_b\phi_c.}
$$
An 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:
$$
\dot Q_a=\int d^3x\,\epsilon_{abc}
[(\nabla^2\phi_b-\lambda\phi_b)\phi_c+\dot\phi_b\dot\phi_c]
=\int_{\partial V}dS\,\epsilon_{abc}(\partial_n\phi_b)\phi_c.
$$
The <mass> term and the products of equal-type derivatives cancel by antisymmetry; <integration by parts> produces the boundary term. \b[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.
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