Let $l=1$, then $\{ x^{\alpha} : |\alpha| = l\} = \{ x_1, x_2, x_3 \}$. Obviously, this is a basis of ${\cal P}_1$, so we only have to prove that $\langle x_i,x_j \rangle = 0$ for all $i,j\in \{ 1,2,3 \}$ with $i \neq j$. So let $i\neq j$ and let $k\in \{1,2,3\}$ be the missing index such that $\{i,j,k\} = \{1,2,3\}$. Then by Fubini's theorem we have $$ \langle x_i,x_j \rangle = \frac{1}{c_1}\int_{\R^3} x_ix_j e^{-\frac 12 (x_i^2+x_j^2+x_k^2)}\, dx = \frac{1}{c_1}\int_{\R} x_i e^{-\frac 12 x_i^2}\, dx_i \int_{\R} x_je^{-\frac 12 x_j^2}\, dx_j \int_{\R} e^{-\frac 12 x_k^2}\, dx_k. $$ Since $x_i \mapsto x_ie^{-\frac 12 x_i^2}$ is an odd function, the first (and also the second) integral is equal to 0, so we have $\langle x_i,x_j \rangle = 0$ for $i\neq j$.

Now let $l=2$. Then $\{ x^{\alpha} : |\alpha| = l\} = \{ x_1^2, x_2^2, x_3^2, x_1x_2, x_1x_3, x_2x_3 \}$. We will show that $\langle x_1^2,x_2^2 \rangle \neq 0$. We have $$ \langle x_1^2,x_2^2 \rangle = \frac{1}{c_2}\int_{\R^3} x_1^2x_2^2 e^{-\frac 12 (x_1^2+x_2^2+x_3^2)}\, dx. $$ This integral is strictly greater than zero because the integrand is a real-valued, non-negative, continuous function on $\R^3$ that is $\not\equiv 0$.