A basic result from linear algebra tells us that because the determinant of $A$ is $0$, the vectors in $\mathbb{R}^n$ that comprise the columns of $A$ must be linearly dependent. This means that:
\begin{eqnarray*}
\lambda_1\begin{pmatrix}\langle v_1,v_1\rangle\\
\langle v_1,v_2\rangle\\\vdots\\
\langle v_1,v_n\rangle
\end{pmatrix}+
\lambda_2\begin{pmatrix}\langle v_2,v_1\rangle\\
\langle v_2,v_2\rangle\\\vdots\\
\langle v_2,v_n\rangle
\end{pmatrix}+
\cdots+
\lambda_n\begin{pmatrix}\langle v_n,v_1\rangle\\
\langle v_n,v_2\rangle\\\vdots\\
\langle v_n,v_n\rangle
\end{pmatrix}=\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}
\end{eqnarray*}
so that for each $i$, $\lambda_1\langle v_1,v_i\rangle+\ldots+\lambda_n\langle v_n,v_i\rangle=0$. Using the properties of the inner product we get that $\langle \lambda_1v_1+\ldots+\lambda_nv_n,v_i\rangle=0$ for all $i$. This implies that $\lambda_1v_1+\ldots+\lambda_nv_n=0$ as desired.
Why does it follows that $\lambda_1v_1+\ldots+\lambda_nv_n=0$? The reason is because the inner product of $\lambda_1v_1+\ldots+\lambda_nv_n$ with all of its "components" is $0$. More explicitly, because $\langle \lambda_1v_1+\ldots+\lambda_nv_n, v_i\rangle=0$ it follows that $\langle \lambda_1v_1+\ldots+\lambda_nv_n, \lambda_iv_i\rangle=0$. Thus, adding all of these inner product expressions, we find that:
\begin{eqnarray*}
\langle \lambda_1v_1+\ldots+\lambda_nv_n, \lambda_1v_1+\ldots+\lambda_nv_n\rangle=0
\end{eqnarray*}
which is only possible if $\lambda_1v_1+\ldots+\lambda_nv_n=0$.
