Example 6 In the following table, only two of the logarithmic values are incorrect. Please correct them:
\begin{tabular}{|c|c|c|c|c|}
\hline$x$ & 0.021 & 0.27 & 1.5 & 2.8 \\
\hline $\lg x$ & $2 a+b+c-3$ & $6 a-3 b-2$ & $3 a-b+c$ & $1-2 a+2 b-c$ \\
\hline \hline$x$ & 3 & 5 & 6 & 7 \\
\hline $\lg x$ & $2 a-b$ & $a+c$ & $1+a-b-c$ & $2(a+c)$ \\
\hline \hline$x$ & 8 & 9 & 14 & \\
\hline $\lg x$ & $3-3 a-3 c$ & $4 a-2 b$ & $1-a+2 b$ & \\
\hline
\end{tabular}
Analyze the given data, which are interrelated. The 11 values of $x$ only involve the prime factors $2, 3, 5, 7$. Using the properties of logarithms, convert the given 11 logarithmic values into expressions involving $\lg 2=1-a-c$, $\lg 3=2a-b$, and $\lg 7=2b+c$. Also, use the condition that "only two values are incorrect" for analysis.

Solution: Since $\lg 9=2 \lg 3$, and $4a-2b=2(2a-b)$, $\lg 0.27=\lg 27-2=3 \lg 3-2$, and $6a-3b-2=3(2a-b)-2$, these three numbers are either all correct or all incorrect. Given that only two data points are incorrect, these three numbers must all be correct. From $\lg 8=3 \lg 2=3-3a-3c \Rightarrow \lg 2=1-a-c$, and $\lg 6=\lg 2+\lg 3=1-a-c+2a-b=1+a-b-c$, $\lg 5=1-\lg 2=a+c$. These three numbers are either all correct or all incorrect, so these three numbers must all be correct. Since $\lg 3$ and $\lg 5$ have been proven correct, $\lg 1.5=\lg 15-1=\lg 3+\lg 5-1=2a-b+a+c-1=3a-b+c-1$, hence $\lg 1.5$ is incorrect. Since $\lg 0.021=\lg 3+\lg 7-3=2a-b+\lg 7-3=2a+b+c-3 \Rightarrow \lg 7=2b+c$, $\lg 14=\lg 2+\lg 7=1-a-c+\lg 7=1-a+2b \Rightarrow \lg 7=2b+c$, $\lg 2.8=\lg 7+2 \lg 2-1=\lg 7+2-2a-2c-1=1-2a+2b-c \Rightarrow \lg 7=2b+c$, these three numbers are either all correct or all incorrect. Since only one is incorrect, these three numbers must all be correct. Thus, $\lg 7=2b+c$ is correct, and $\lg 7=2(a+c)$ is incorrect. Therefore, the values of $\lg 1.5$ and $\lg 7$ in the table are incorrect, and should be $\lg 1.5=3a-b+c-1$ and $\lg 7=2b+c$.