We have points $A\left(1, x, 3\right), B\left(3 , 4, 7\right) \mathrm{and} C\left(y, -2, -5\right)$, which are collinear.

Now, $\overrightarrow{AB}=2\hat{i}+\left(4-x\right)\hat{j}+4\hat{k}$, $\overrightarrow{BC}=\left(y-3\right)\hat{i}-6\hat{j}-12\hat{k}$ and $\overrightarrow{CA}=\left(1-y\right)\hat{i}+\left(x+2\right)\hat{j}+8\hat{k}$

Then, $\overrightarrow{AB}=m\overrightarrow{CA} \Rightarrow 2\hat{i}+\left(4-x\right)\hat{j}+4\hat{k}=m\left[\left(1-y\right)\hat{i}+\left(x+2\right)\hat{j}+8\hat{k}\right]$

Comparing both sides,$2=m\left(1-y\right), 4-x=m\left(x+3\right) \mathrm{and} 4=8m$

$\Rightarrow m=\frac{1}{2}=\frac{4-x}{x+2}=\frac{2}{1-y}$

Solving, we get $x=2 \mathrm{and} \mathrm{y}=-3$

Hence, $x+y=-1$