<s>[INST] Making vector subject of equation I have the following question

Solve the vector equation $\bf{a \times r}\ +\ \lambda\bf{r = c}$ for
r where $\lambda \ne 0$.

So far my approach has been

 Dot both sides with c, which with some rearranging leads to $\bf{r}.(\bf{c \times a}\ +\ \lambda \bf{c}) = |{\bf{c}}|^2$. Leading me to believe that r is a point on a plane. I then converted this equation to Cartesian form and then to Parametric for so r is the subject.

However my answer looks very different to the provided answer and no working is provided.
The solution given was

 $$\bf{r} = N^{-1} \left(\bf{(a.c)a} + \lambda^2\bf{c} - \lambda \bf{a} \times \bf{c}\right)\ \text{where}\ N = \lambda(k^2 + \lambda^2)$$

Is my answer equivalent to this and how was this solution reached?
 [/INST] First off, I assume that in the provided solution, $\mathbf{k}$ should be $\mathbf{a}$, since the former doesn't appear in the problem statement, and after swapping it for $\mathbf{a}$ the answer coincides with mine.
The provided answer contains the term $\mathbf{a}\times\mathbf{c}$, which gives us a hint that we need to apply $\mathbf{a}\times$ to the equation. Indeed doing so gives us
$$(\mathbf{a}\cdot\mathbf{r})\mathbf{a} - (\mathbf{a}\cdot\mathbf{a})\mathbf{r} + \lambda\mathbf{a}\times\mathbf{r} = \mathbf{a}\times \mathbf{c}\ .$$
Now observe that we can use the original equation to substitute $\mathbf{a}\times\mathbf{r}$ with $\mathbf{c} -\lambda \mathbf{r}$, so that $(\mathbf{a}\cdot \mathbf{r})\mathbf{a} - |\mathbf{a}|^2 \mathbf{r} + \lambda\mathbf{c} - \lambda^2 \mathbf{r} = \mathbf{a}\times \mathbf{c}$, or
$$
(\mathbf{a}\cdot\mathbf{r})\mathbf{a}-(|\mathbf{a}|^2+\lambda^2)\mathbf{r} = \mathbf{a}\times\mathbf{c} - \lambda\mathbf{c}\ .
$$
At this point we still have a pesky $\mathbf{a}\cdot\mathbf{r}$ factor remaining, but since it's the coefficient of $\mathbf{a}$, it gets annihilated if we take the vector product with $\mathbf{a}$ again,
$$
-(|\mathbf{a}|^2 + \lambda^2)\mathbf{a}\times\mathbf{r} = (\mathbf{a}\cdot \mathbf{c})\mathbf{a} - (\mathbf{a}\cdot\mathbf{a})\mathbf{c} - \lambda\mathbf{a}\times \mathbf{c}\ .
$$
This might alarm you, because we seem to be reintroducing $\mathbf{a}\times \mathbf{r}$, but we already know how to transform it into something simpler. Making the same substitution for $\mathbf{a}\times\mathbf{r}$ as before, we get
$$
-(|\mathbf{a}|^2 + \lambda^2)(\mathbf{c} - \lambda\mathbf{r}) = (\mathbf{a}\cdot \mathbf{c})\mathbf{a} - |\mathbf{a}|^2\mathbf{c} - \lambda\mathbf{a}\times \mathbf{c}\ .
$$
Finally the $-|\mathbf{a}|^2\mathbf{c}$ terms will cancel, and rearranging the equation gives us the provided answer.
There are two main ideas behind the solution. The first is to use the properties of the cross product and vector triple product to get rid of factors that involve products of the unknown $\mathbf{r}$ with other vectors, so that we end up with an equation containing only a linear factor of $\mathbf{r}$ with a constant coefficient. The second is to use the original equation to substitute $\mathbf{a}\times\mathbf{r}$ repeatedly by an expression where $\mathbf{r}$ is only multiplied by a constant scalar.
Your attempt sounds correct in principle, but perhaps a little too "coordinate-y". In essence, reversing dot products is not a particularly easy thing to do. However, at least the expression $\mathbf{r}\cdot(\mathbf{c}\times\mathbf{a} + \lambda\mathbf{c}) = |\mathbf{c}|^2$ is consistent with the solution. To see this, take the dot product of the provided solution with $\mathbf{c}\times\mathbf{a} + \lambda\mathbf{c}$. In the denominator, all the triple products of $\mathbf{a}$, $\mathbf{a}$, $\mathbf{c}$ or of $\mathbf{a}$, $\mathbf{c}$, $\mathbf{c}$ vanish, so what remains is
$$
\lambda|\mathbf{a}\cdot\mathbf{c}|^2 + \lambda^3|\mathbf{c}|^2 + \lambda|\mathbf{a}\times\mathbf{c}|^2\ .
$$
Here the term $|\mathbf{a}\times\mathbf{c}|^2$ can be rewritten as a scalar triple product $(\mathbf{a}\times\mathbf{c})\cdot(\mathbf{a}\times\mathbf{c})$. We can cyclically permute it into $\mathbf{a}\cdot(\mathbf{c}\times(\mathbf{a}\times\mathbf{c}))$, and expand the result using the vector triple product relation, i.e. $\mathbf{a}\cdot[(\mathbf{c}\cdot\mathbf{c})\mathbf{a} - (\mathbf{c}\cdot\mathbf{a})\mathbf{c}] = |\mathbf{a}|^2|\mathbf{c}|^2 - |\mathbf{a}\cdot\mathbf{c}|^2$. Notice that second term here leads to a cancellation, so we end up with
$$
\lambda|\mathbf{a}|^2|\mathbf{c}|^2 + \lambda^3|\mathbf{c}|^2\ .
$$
After dividing by $N = \lambda(|\mathbf{a}|^2 + \lambda^2)$, you get $|\mathbf{c}|^2$, which is the right hand side of the equation you obtained.
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