In an elementary linear algebra (or calculus) course, we would solve this problem as follows. First, we would need two vectors parallel to the plane. If \(\bbm x\\y\\z\ebm\) lies in the plane, then \(x-2y+4z=0\text{,}\) so \(x=2y-4z\text{,}\) and
\begin{equation*}
\bbm x\\y\\z\ebm = \bbm 2y-4z\\y\\z\ebm = y\bbm 2\\1\\0\ebm + z\bbm -4\\0\\1\ebm\text{,}
\end{equation*}
so \(\uu=\bbm 2\\1\\0\ebm\) and \(\vv\bbm -4\\0\\1\ebm\) are parallel to the plane. We then compute the normal vector
\begin{equation*}
\mathbf{n}=\uu\times\vv=\bbm 1\\-2\\4\ebm\text{,}
\end{equation*}
and compute the projection of the position vector \(\mathbf{p}=\bbm 3,1,-2\ebm\) for the point \(P=(3,1,-2)\) onto \(\mathbf{n}\text{.}\) This gives the vector
\begin{equation*}
\xx = \left(\frac{\mathbf{p}\dotp\mathbf{n}}{\len{\mathbf{n}}^2}\right)\mathbf{n} = \frac{-7}{21}\bbm 1\\-2\\4\ebm =\bbm-1/3\\2/3\\-4/3\ebm\text{.}
\end{equation*}
Now, this vector is parallel to \(\mathbf{n}\text{,}\) so itβs perpendicular to the plane. Subtracting it from \(\mathbf{p}\) gives a vector parallel to the plane, and this is the position vector for the point we seek.
\begin{equation*}
\mathbf{q}=\mathbf{p}-\xx=\bbm 3\\1\\-2\ebm-\bbm -1/3\\-2/3\\-4/3\ebm = \bbm 10/3\\1/3\\-2/3\ebm
\end{equation*}
so the closest point is \(Q=\bigl(\frac{10}{3},\frac13,-\frac{2}{3}\bigr)\text{.}\) We werenβt asked for it, but note that if we wanted the distance from the point \(P\) to the plane, this is given by \(\len{\xx}=\frac13\sqrt{21}\text{.}\)