Prove that the point A $\displaystyle (i+2j-3k)$ lies on the line L1 $\displaystyle

r = \left(\begin{array}{c}4\\-4\\3\end{array}\right) + \mu \left(\begin{array}{c}1\\-2\\2\end{array}\right)

$

So what i have done is:

$\displaystyle \left(\begin{array}{c}1\\2\\-3\end{array}\right) = \left(\begin{array}{c}4\\-4\\3\end{array}\right) + \mu \left(\begin{array}{c}1\\-2\\2\end{array}\right)$

$\displaystyle \left(\begin{array}{c}1\\2\\-3\end{array}\right) - \left(\begin{array}{c}4\\-4\\3\end{array}\right) = \mu \left(\begin{array}{c}1\\-2\\2\end{array}\right)$

$\displaystyle \left(\begin{array}{c}-3\\6\\-6\end{array}\right) = \mu \left(\begin{array}{c}1\\-2\\2\end{array}\right)$

Therefore: $\displaystyle \mu = -3$

so Point A deos lie on the line L1.

am not sure if this is the correct way to prove it.