Given vectors are $\vec{a}=p \hat{i}-2 \hat{j}+5 \hat{k}$ and $\vec{b}=p \hat{i}+q \hat{j}-3 \hat{k}$ According to questions, vectors are collineur.
$$
\begin{aligned}
& \overrightarrow{\mathrm{a}}=\lambda \overrightarrow{\mathrm{b}} \\
& (\hat{\mathrm{P}}-2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})=\lambda(\hat{\mathrm{i}}+\mathrm{q} \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \\
& \mathrm{P} \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}=\lambda \hat{\mathrm{i}}+\mathrm{q} \lambda \hat{\mathrm{j}}-3 \lambda \hat{\mathrm{k}} \\
& \hat{\mathrm{i}}(\mathrm{p}-\lambda)+\hat{\mathrm{j}}(-2+\mathrm{q} \lambda)+\hat{\mathrm{k}}(5+3 \lambda)=0
\end{aligned}
$$
compare each comonent,
$$
\begin{aligned}
& 5+3 \lambda=0 \\
& \lambda=\frac{-5}{3} \\
& \Rightarrow \mathrm{P}=\lambda \\
& \mathrm{P}=\frac{-5}{3} \\
& \Rightarrow-\mathrm{q} \lambda-2=0 \\
& \mathrm{q}=\frac{-2}{\lambda}=\frac{-2 \times 3}{-5}=\frac{6}{5}
\end{aligned}
$$
So, $\mathrm{p}=\frac{-5}{3}, \mathrm{q}=\frac{6}{5}$