Given vectors: $\overrightarrow{a}=\overrightarrow{p}+2\overrightarrow{q},\overrightarrow{b}=3\overrightarrow{p}-\overrightarrow{q}$ where $|\overrightarrow{p}|=2,|\overrightarrow{q}|=6,\angle(\overrightarrow{p},\overrightarrow{q})=\pi/3$. Find vector of triangle height constructed over vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ that is orthogonal to vector $\overrightarrow{a}$.
Let the vector of triangle height is $\overrightarrow{h}$. Since $\overrightarrow{h}$ is orthogonal to $\overrightarrow{a}$, we can look at right-angled triangle formed of vectors $\overrightarrow{h},\overrightarrow{x_1},\overrightarrow{x}$ where $\overrightarrow{x_1}$ is the component of $\overrightarrow{x}$ onto $\overrightarrow{a}$ and $\overrightarrow{x}$ is hypotenuse.
Orthogonal projection of $\overrightarrow{x}$ onto $\overrightarrow{a}$ is given by $$\overrightarrow{x_1}=proj_{\overrightarrow{a}} {\overrightarrow{x}}=\frac{\overrightarrow{a}\cdot \overrightarrow{x}}{|\overrightarrow{a}|^2}\cdot \overrightarrow{a}$$
But we don't know the coordinates or magnitude of $\overrightarrow{a}$ and $\overrightarrow{x}$.
Question: How to find vector $\overrightarrow{h}$?
Dot product of $\overrightarrow{p},\overrightarrow{q}$ is $|\overrightarrow{p}\cdot \overrightarrow{q}|=|\overrightarrow{p}||\overrightarrow{q}|\cos\angle(\overrightarrow{p},\overrightarrow{q})=6$.
Cross product of $\overrightarrow{p},\overrightarrow{q}$ is $|\overrightarrow{p}\times \overrightarrow{q}|=|\overrightarrow{p}||\overrightarrow{q}|\sin\angle(\overrightarrow{p},\overrightarrow{q})=6\sqrt 3$.
Cross product of $\overrightarrow{a}$ and $\overrightarrow{b}$ is $|\overrightarrow{a}\times \overrightarrow{b}|=6(6\sqrt 3-11)$.
HINTS:
You write of vector $\vec x$, but you really mean vector $\vec b$, at least if the question (which is badly worded) makes any sense.
You have enough information to find $|\vec a|$, $|\vec b|$, and $\vec a\cdot\vec b$. For example,
$$\begin{align} |\vec a| &= \sqrt{\vec a\cdot\vec a} \\[2ex] &= \sqrt{(\vec p+2\vec q)\cdot(\vec p+2\vec q)} \\[2ex] &= \sqrt{\vec p\cdot\vec p+4\vec p\cdot\vec q+4\vec q\cdot\vec q} \\[2ex] &= \sqrt{|\vec p|^2+4(\vec p\cdot\vec q)+4|\vec q|^2} \\[2ex] &= \sqrt{2^2+4(6)+4(6^2)} \\[2ex] &= \sqrt{172} \\[2ex] &= 2\sqrt{43} \end{align}$$
Then use those values to find $\vec{x_1}$ and thus $\vec h$.