Solve for $\vec{r}$ where
\begin{align} \vec{r}.\vec{n}_1&=1\\ \vec{r}.\vec{n}_2&=1\\ \vec{n}_1&\not=\vec{n}_2 \end{align}
I took
$$\vec{r}=a.\vec{n}_1+b.\vec{n}_2+c.(\vec{n}_1x\vec{n}_2)$$
then
\begin{align}
\vec{r}.\vec{n}_1=a.\vec{n}_1.\vec{n}_1+b.\vec{n}_2.\vec{n}_1&=1\\
\vec{r}.\vec{n}_2=a.\vec{n}_1.\vec{n}_2+b.\vec{n}_2.\vec{n}_2=1
\end{align}
Subtracting we get
$$(a.\vec{n}_1+b.\vec{n}_2).(\vec{n}_1-\vec{n}_2)=0 $$
I have no idea what to do next.
https://i.stack.imgur.com/zOM6s.jpg
You certainly need that $\vec{n}_1$ and $\vec{n}_2$ are not multiples of each other (or one is not a multiple of the other, to be more precise). If, say, $\vec{n}_2=\lambda\vec{n}_1$, then $\vec{r}.\vec{n}_2=\lambda\vec{r}.\vec{n}_1$ and they cannot both be $1$ unless $\lambda=1$.
Suppose now that neither of $\vec{n}_1$ and $\vec{n}_2$ is a multiple of the other. Then, consider $$ c_1(\vec{n}_1-\operatorname{proj}_{\vec{n}_2}\vec{n}_1)+c_2(\vec{n}_2-\operatorname{proj}_{\vec{n}_1}\vec{n}_2). $$ We observe that if we take the dot product with $\vec{n}_1$, the second term disappears since we are subtracting the portion of $\vec{n}_2$ in the direction of $\vec{n}_1$. Therefore, we are left with $$ c_1\vec{n}_1.(\vec{n}_1-\operatorname{proj}_{\vec{n}_2}\vec{n}_1). $$ Now, you want this to be $1$, which can happen by a good choice of $c_1$ unless $$ \vec{n}_1.(\vec{n}_1-\operatorname{proj}_{\vec{n}_2}\vec{n}_1)=0. $$ Multiplying this out, we get $$ |\vec{n}_1|^2-\frac{(\vec{n}_1.\vec{n}_2)^2}{|\vec{n}_2|^2}=0, $$ or that $$ |\vec{n}_1|^2|\vec{n}_2|^2=(\vec{n}_1.\vec{n}_2)^2. $$ Since $\vec{n}_1.\vec{n}_2=|\vec{n}_1||\vec{n}_2|\cos(\theta)$ where $\theta$ is the angle between the vectors, substituting this in implies that $$ \cos^2(\theta)=1 $$ or that $\theta=0$ or $\theta=\pi$. Neither case is possible because then one vector is a multiple of the other. We have a similar case when dealing with $c_2$. Therefore, $$ \vec{r}=\frac{\vec{n}_1-\operatorname{proj}_{\vec{n}_2}\vec{n}_1}{\vec{n}_1.(\vec{n}_1-\operatorname{proj}_{\vec{n}_2}\vec{n}_1)}+\frac{\vec{n}_2-\operatorname{proj}_{\vec{n}_1}\vec{n}_2}{\vec{n}_2.(\vec{n}_2-\operatorname{proj}_{\vec{n}_1}\vec{n}_2)}. $$