How to find vector from point along line lying on a direction vector?

Sorry if the title's confusing. I've been trying to solve this problem with no avail:

Let $V = (0,3,2)$ and $A = (-1,0,1).$ Let $P$ be the point on the line passing through $A$ with direction vector $(1,1,1)$ that is closest to $V.$ Find $\overrightarrow{AP}.$

Obviously, to solve the problem I need to find the point P. However, I'm quite lost in regards to how to find it. I'm pretty sure I can use the formula to find the distance between the line (and P) from point V, but I don't know how I would find the point P from the distance. I'm also confused with the specific wording of the problem. I'm pretty sure that it dictates the line lying along the direction vector $(1, 1, 1)$ is the same line that P is on, but I'm not sure. Any thoughts or solutions are appreciated in advance!

In this case we can use the projection formula $$\operatorname{proj}_{b}a=\frac {a\cdot b}{\|b\|^2}b.$$

Let $L_A=\{(-1,0,1)+t\,(1,1,1)\,|\,t\in \mathbb{R}\}$ be the line passing through $A$ with direction vector $(1,1,1)$

Also, $\vec{AV}=(0,3,2)-(-1,0,1)=(1,3,1)$.

So, $$\vec{AP}=\operatorname{proj}_{(1,1,1)}\vec{AV}=\frac {\vec{AV}\cdot (1,1,1)}{\|(1,1,1)\|^2}(1,1,1)$$ $$=\frac{(1,3,1)\cdot (1,1,1)}{3}(1,1,1)$$ $$\Rightarrow \vec{AP}=\frac{5}{3}(1,1,1).$$

One can deduce from this information, that $P=(-1+a, a,1+a)$ for some $a\in\mathbb R$. The distance $|PV|$ is then given by $$f(a) = \sqrt{ (a-1)^2 +(a-3)^2 + (a-1)^2} $$ To minimize this, square and take the derivative and set it equal to zero: $$ \left( 2(a-1)^2 +(a-3)^2 \right)’ =0 \\ 4(a-1) +2(a-3) = 0 \\ a=\frac 53 \\ \therefore P=\left(\frac 23, \frac 53, \frac 83\right)$$ and $$\vec{AP} =\bigg\langle \frac 53, \frac 53, \frac 53 \bigg\rangle$$