<s>[INST] How to find the position vector for the point of intersection of a line and the perpendicular line through a point C 
How could I find the exact coordinates of the point N for example which is the point of intersection of the line $L=\begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix} +t\begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}$ and a perpendicular line that passes through a point C(1,2,3) not on the line.

What I have done so far is find the value of point N in terms of $t$ to find the vector $\overrightarrow{CN}$:
$$\longrightarrow N\begin{pmatrix} -3t \\ -2t-1 \\ -3t+2 \end{pmatrix}$$
$$\longrightarrow \overrightarrow{CN}=\begin{pmatrix} -3t \\ -2t-1 \\ -3t+2 \end{pmatrix}-\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$$
$$=\begin{pmatrix} -3t-1 \\ -2t-3 \\ -3t-1 \end{pmatrix}$$
After this I made the scalar product eual zero:
$$\overrightarrow{CN} \cdot \begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}=0$$
With this I find that, $$t=\frac{-6}{11}$$
But my textbook says that $$t=\frac{5}{11}$$
Can anyone confirm which answer for $t$ is right and if it is $t=\frac{5}{11}$ then what did I do wrong?
 [/INST] Your derivation seems correct indeed, by a direct check, for $t=-\frac 6{11}$ we obtain
$$N=\begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix} -\frac 6{11}\begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}=\begin{pmatrix} \frac {18}{11} \\ \frac {1}{11} \\ \frac {40}{11} \end{pmatrix}$$
then
$$\overrightarrow{CN}=\begin{pmatrix} \frac {7}{11} \\ -\frac {21}{11} \\ \frac {7}{11} \end{pmatrix}$$
and
$$\begin{pmatrix} \frac {7}{11} \\ -\frac {21}{11} \\ \frac {7}{11} \end{pmatrix} \cdot \begin{pmatrix} -3 \\ -2 \\ -3 \end{pmatrix}= -\frac {21}{11}+\frac {42}{11}-\frac {21}{11}=0$$
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