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> Referred to the origin $O$, the points $P$ and $Q$ have position vectors **i** + $2$**j** + $3$**k** and $3$**i** + $3$**j** + $5$**k** respectively. Find the position vector of the point where the line $PQ$ meets the plane $y=0$
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I’ve found the direction of line $PQ$:

$\overrightarrow{PQ}$

$= \overrightarrow {OQ} - \overrightarrow {OP}$

$=$ $3$**i** + $3$**j** + $5$**k** - ( **i** + $2$**j** + $3$**k**)

$=$ $2$**i** + **j** + $2$**k**

So, the vector equation of line $PQ$ is **r** = **i** + $2$**j** + $3$**k** + $\lambda$($2$**i** + **j** + $2$**k**)

To find the position vector of the point of intersection between line $PQ$ and the plane $y=0$, I understand that I need to substitute **r** = **i** + $2$**j** + $3$**k** + $\lambda$($2$**i** + **j** + $2$**k**) into the equation of plane, **r**$\cdot$**n** = $d$, where **n** is a vector perpendicular to the plane and $d$ is a constant.

Is it right to assume that **n**= $\left(\begin{matrix}2\\1\\2\\\end{matrix}\right)$ ?

How do I find the value of $d$? How do I apply the information of plane $y=0$?

Any hints would be helpful. Thank you