Given that
$\begin{aligned}
\mathbf{A} & =\hat{\mathbf{i}}+x \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{B}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}, \\
\mathbf{C} & =y \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \\
\mathbf{A B} & =2 \hat{\mathbf{i}}+(4-x) \hat{\mathbf{j}}+4 \hat{\mathbf{k}},
\end{aligned}$
and $\mathbf{B C}=(y-3) \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-12 \hat{\mathbf{k}}$
$\Rightarrow$ Since, A, B, C are collinear, then
$\begin{aligned}
& \mathbf{A B}=t \mathrm{BC} \\
& 2 \hat{\mathbf{i}}+(4-x) \hat{\mathbf{j}}+4 \hat{\mathbf{k}}=t\{(y-3) \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-12 \hat{\mathbf{k}}\} \\
& \Rightarrow 2 \hat{\mathbf{i}}+(4-x) \hat{\mathbf{j}}+4 \hat{\mathbf{k}}=t(y-3) \hat{\mathbf{i}}-6 t \hat{\mathbf{j}}-12 t \hat{\mathbf{k}}
\end{aligned}$
Equating the coefficient, of $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$
$\begin{aligned}
t(y-3) & =2, \\
4-x & =-6 t, \text { and } 4=-12 t \\
t & =-\frac{1}{3}
\end{aligned}$
$\begin{aligned} & \therefore 4-x=-6\left(-\frac{1}{3}\right)=2 \Rightarrow x=2 \\ & \text { and } \quad-\frac{1}{3}(y-3)=2 \\ & y-3=-6 \Rightarrow y=-3 \\ & \text { Then }(x, y)=(2,-3) \\ & \end{aligned}$