If the points \(\mathrm{P}(4,5, \mathrm{x}), \mathrm{Q}(3, \mathrm{y}, 4)\) and \(\mathrm{R}(5,8,0)\) are collinear, then the value of \(x+y\) is

\(
\begin{aligned}
& \overline{\mathrm{PQ}}=-\hat{\mathrm{i}}+(\mathrm{y}-5) \hat{\mathrm{j}}+(4-\mathrm{x}) \hat{\mathrm{k}} \\
& \overline{\mathrm{PR}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}=\mathrm{x} \hat{\mathrm{k}}
\end{aligned}
\)
Since points \(\mathrm{P}\). Q, R are collinear
\(\overline{\mathrm{PQ}}=\mathrm{a} \overline{\mathrm{PR}} \)
\( \therefore-\hat{\mathrm{i}}+(\mathrm{y}-5) \hat{\mathrm{j}}+(4-\mathrm{x}) \hat{\mathrm{k}}=\mathrm{a}(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}=\mathrm{x} \hat{\mathrm{k}}) \)
\( \therefore \mathrm{a}=-1,3 \mathrm{a}=\mathrm{y}-5,-\mathrm{ax}=4-\mathrm{x} \)
\( \therefore \mathrm{a}=-1 \Rightarrow-3=\mathrm{y}-5 \quad \text { i.e. } \mathrm{y}=2 \text { and } \mathrm{x}=4\) \(-\mathrm{x} \Rightarrow \mathrm{x}=2 \Rightarrow \)
\( \mathrm{x}+\mathrm{y}=4\)