### Instruction: Find the point on the line defined by
\[\begin{pmatrix} 4 \\ 0 \\ 1 \end{pmatrix} + t \begin{pmatrix} -2 \\ 6 \\ -3 \end{pmatrix}\]that is closest to the point $(2,3,4).$.### Response: To find the closest point on the line, I need to minimize the distance between the point $(2,3,4)$ and the line.
The distance between a point and a line is the length of the perpendicular segment that connects them.
So I need to find the value of $t$ that makes the segment from $(2,3,4)$ to the line perpendicular to the direction of the line.
The direction of the line is given by the vector $\begin{pmatrix} -2 \\ 6 \\ -3 \end{pmatrix}$.
So I need to find $t$ such that the dot product of this vector and the segment from $(2,3,4)$ to the line is zero.
The segment from $(2,3,4)$ to the line is given by $\begin{pmatrix} 4 \\ 0 \\ 1 \end{pmatrix} + t \begin{pmatrix} -2 \\ 6 \\ -3 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 2-2t \\ -3+6t \\ -3-3t \end{pmatrix}$.
So I need to solve the equation $\begin{pmatrix} -2 \\ 6 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 2-2t \\ -3+6t \\ -3-3t \end{pmatrix} = 0$.
Expanding the dot product, I get $-4+4t-18+36t+9+9t=0$.
Simplifying, I get $49t-13=0$.
Solving for $t$, I get $t=\frac{13}{49}$.
To find the point on the line, I need to plug this value of $t$ into the parametric equation of the line.
The point is $\begin{pmatrix} 4 \\ 0 \\ 1 \end{pmatrix} + \frac{13}{49} \begin{pmatrix} -2 \\ 6 \\ -3 \end{pmatrix} = \begin{pmatrix} 4-\frac{26}{49} \\ \frac{78}{49} \\ 1-\frac{39}{49} \end{pmatrix} = \begin{pmatrix} \frac{170}{49} \\ \frac{78}{49} \\ \frac{10}{49} \end{pmatrix}$.