Given $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}, \overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-7 \hat{\mathrm{k}}$
Here, $\vec{r} \cdot \vec{a}=9$. and $(2 \hat{i}+3 \hat{j}+\hat{k})=9$
Similarly, $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{b}}=7$ and $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{c}}=6$
$$
\begin{aligned}
& 4 x+y=7 ...(ii)\\
& x-3 y-7 z=6...(iii)
\end{aligned}
$$
Multiply eq. (i) by 7
$$
14 x+21 y+7 z=63...(iv)
$$
Add (iii) and (iv)
$$
\begin{aligned}
& 14 x+21 y+7 z=63 \\
& x-3 y-7 z=6 \\
& \hline 15 x+18 y=69 \\
& 5 x+6 y=23...(v)
\end{aligned}
$$
Multiply equation (ii) by $b$.
$$
\begin{aligned}
24 x+6 y & =42 \\
5 x \pm 6 y & =23 \\
\hline 19 x & =19
\end{aligned}
$$
$$
x=1
$$
From (ii),
$$
\begin{gathered}
4 x+y=7 \\
4+y=7 \\
y=3
\end{gathered}
$$
From (i)
$$
\begin{aligned}
& 2 \mathrm{x}+3 \mathrm{y}+\mathrm{z}: 9 \Rightarrow 2 \times 1+3 \times 3+2=9 \\
& \Rightarrow \mathrm{z}=9-2-9=-2
\end{aligned}
$$
So, $(x, y, z) \rightarrow(1,3,-2)$.