Let, $\vec{c}=a \hat{i}+b \hat{j}+c \hat{k}$
Given, $\vec{c} \times(\hat{i}+2 \hat{j}+5 \hat{k})=\overrightarrow{0}$
$$
\begin{aligned}
&\Rightarrow\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a & b & c \\
1 & 2 & 5
\end{array}\right|=\overrightarrow{0} \\
&\Rightarrow(5 b-2 c) \hat{i}-(5 a-c) \hat{j}+(2 a-b) \hat{k} \\
&=0 \hat{i}+0 \hat{j}+0 \hat{k}
\end{aligned}
$$
Comparing both sides, we get $5 b-2 c=0 ; 5 a-c=0 ; 2 a-b=0$ or $5 b=2 c ; 5 a=c ; 2 a=b$
Also given $|\vec{c}|^2=60$
$$
\Rightarrow a^2+b^2+c^2=60
$$
Putting the value of $b$ and $c$ in above eqn., we get

$$
\begin{aligned}
&a^2+(2 a)^2+(5 a)^2=60 \\
&\Rightarrow a^2+4 a^2+25 a^2=60 \Rightarrow 30 a^2=60 \\
&a^2=2 \\
&a=\pm \sqrt{2} ; b=2 \sqrt{2} ; c=5 \sqrt{2}
\end{aligned}
$$
Now, $\vec{c}=a \hat{i}+b \hat{j}+c \hat{k}$
$$
\therefore \vec{c}=\sqrt{2} \hat{i}+2 \sqrt{2} \hat{j}+5 \sqrt{2} \hat{k}
$$
Value of $\vec{c} \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is
$$
\begin{aligned}
&(\sqrt{2} \hat{i}+2 \sqrt{2} \hat{j}+5 \sqrt{2} \hat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \\
&=-7 \sqrt{2}+4 \sqrt{2}+15 \sqrt{2}=12 \sqrt{2}
\end{aligned}
$$