Hello stealthmaths

OK, here's the working using the other notation.

The line $\displaystyle Q$ has equation:$\displaystyle \textbf{r} = 2\textbf i + \textbf j+3\textbf k+\lambda(\textbf i + 3\textbf j-5\textbf k)$

This means that, for any value of $\displaystyle \lambda$, the point $\displaystyle R$ with position vector $\displaystyle \textbf r$ is any general point on the line $\displaystyle Q$.

The line joining the point $\displaystyle P$ to the point $\displaystyle R$ has vector given by:$\displaystyle \textbf{PR} = \lambda\textbf i + (-2+3\lambda)\textbf j+(4-5\lambda)\textbf k$

The vector that determines the direction of $\displaystyle Q$, $\displaystyle \textbf q$, say, is:$\displaystyle \textbf q =\textbf i + 3\textbf j-5\textbf k$

So PR is perpendicular to $\displaystyle Q$ if:$\displaystyle \textbf q.\textbf{PR} = 0$

$\displaystyle \Rightarrow (\textbf i + 3\textbf j-5\textbf k).\Big(\lambda\textbf i + (-2+3\lambda)\textbf j+(4-5\lambda)\textbf k\Big) = 0$

$\displaystyle \Rightarrow \lambda+3(-2+3\lambda)-5(4-5\lambda)=0$

Solve for $\displaystyle \lambda$ and substitute back into the equation for $\displaystyle \textbf r$ to find the coordinates of $\displaystyle R$.

Can you complete it now?

Grandad