Check if two vector equations of parametric surfaces are equivalent Give the vector equation of the plane through these lines:
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}\,\,\,$ and $\,\,\,\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\0\\3\end{pmatrix}+\mu\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}$.
My answer is: $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\-1\\2\end{pmatrix}+\mu\cdot\begin{pmatrix}0\\-3\\1\end{pmatrix}$. The solutions manual suggests the following equation, which is the equation of a straight line (probably a typo): $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}+\mu\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}$
Is my solution the right solution? Could someone provide a general way to check?

You have two planes that go through the same point (4,1,1). Now all you need is having them parallel. You can check this by forming the cross product of the defining vectors, i.e. (0,-1,2) and (0,-3-1) for the first plane, that will give you a normal to the plane. Do the same for the second plane; if your normal vectors are parallel, so are the planes and hence identical in this case.
