Finding the point of intersection of line 

$\overrightarrow{r}=\overrightarrow{a}+2\overrightarrow{b}+p\left(\overrightarrow{a}-2\overrightarrow{c}\right) ...\left(1\right)$ and $\overrightarrow{r}=3\overrightarrow{a}-q\left(\overrightarrow{c}-\overrightarrow{b}\right)+k\left(\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\right) ...\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$

$\overrightarrow{a}+2\overrightarrow{b}+p\left(\overrightarrow{a}-2\overrightarrow{c}\right) =3\overrightarrow{a}-q\left(\overrightarrow{c}-\overrightarrow{b}\right)+k\left(\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\right)$

$\Rightarrow \overrightarrow{a}\left(1+p\right)+2\overrightarrow{b}+\left(-2p\right)\overrightarrow{c} =\overrightarrow{a}\left(3+k\right)+\overrightarrow{b}\left(q-k\right)+\overrightarrow{c}\left(-q+k\right)$

Comparing LHS and RHS,

$1+p=3+k\Rightarrow p-k=2 , q-k=2 \mathrm{and} -q+k=-2p$

Solving above equations for $p, q \mathrm{and} k$, we get $p=1, q=1 \mathrm{and} \mathrm{k}=-1$

In order to find the point of intersection of given line and plane,

$\overrightarrow{r}=\overrightarrow{a}+2\overrightarrow{b}+p\left(\overrightarrow{a}-2\overrightarrow{c}\right)=2\overrightarrow{a}+2\overrightarrow{b}-2\overrightarrow{c}$

As, the point of intersection given is 

$\overrightarrow{r}=x\overrightarrow{a}+y\overrightarrow{b}-z\overrightarrow{c}$

So, $xyz=2\times 2\times \left(-2\right)=-8$