Finding $P$ knowing $\overrightarrow{PQ}×\overrightarrow{b}$, $\overrightarrow{PQ}⋅\overrightarrow{c}$, $\overrightarrow{b}$, and $\overrightarrow{c}$

Let $Q$ be the point $(1,2,3)$, let $\overrightarrow{b} = \langle -1, 0, 1\rangle$, and let $\overrightarrow{c} = \langle 2, 1, 5\rangle$. It is known that $\overrightarrow{PQ} \times \overrightarrow{b} = \overrightarrow{0}$ and that $\overrightarrow{PQ} \cdot \overrightarrow{c} = 5$.

Find the point $P$.

I have no idea where to begin to solve this problem, but I do know the generic forms of the dot product and cross product, as so:

Generic equation for the cross product:

$$\overrightarrow{a} \times \overrightarrow{b} =(a_2 b_3 - a_3 b_2)\overrightarrow{i} + (a_3 b_1 - a_1 b_3)\overrightarrow{j} + (a_1 b_2 - a_2 b_1)\overrightarrow{k}$$

Generic equation for the dot product:

$$\overrightarrow{a} \cdot \overrightarrow{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$$

So, using this knowledge we can derive an expression for $\overrightarrow{PQ}$, given the representation of point $P$ as $(p_1, p_2, p_3)$:

$$\overrightarrow{PQ} = \langle 1-p_1, 2-p_2, 3-p_3 \rangle$$

Then, in order to satisfy $\overrightarrow{PQ} \cdot \overrightarrow{c} = 5$, we can write the expression:

$$\langle 2, 1, 5 \rangle \cdot \langle 1-p_1, 2-p_2, 3-p_3 \rangle = (2-2p_1) + (2-p_2) + (15-5p_3) = 5$$

But I'm not sure where to go from here. I assume I could solve for a generic vector in terms of $P$ and combine this function with the one using the cross product, in a similar fashion.

How can I find the value of point $P$ by combining the dot product and cross product formulas?

Thank you so much for your help; it is super appreciated!

If we consider the dot product result (which you provided), and cross product equation (which returns the result $\overrightarrow{PQ}×\overrightarrow{b}=<-p_1,2p_2,-p_3>$), we get a system of equations:

$$2p_1 + 2p_2 + 5p_3 = 14 \\ -p_1+2p_2-p_3=0$$

after much algebraic manipulation, we get: $p_1=\frac{14}{3}, p_2=\frac{14}{6}, p_3=0$

Hint.

$\overrightarrow{PQ} \times \overrightarrow{b} = \overrightarrow{0}$ iff $P$ lies on the line passing through $Q$ and parallel to $\overrightarrow{b}$.

$\overrightarrow{PQ} \cdot \overrightarrow{c} = 5$ is the equation of a plane.

At the end $P$ is the intersection point of the line and the plane.