6-10 In the following table, two of the logarithmic values are incorrect. Please correct them.
\begin{tabular}{|c|c|c|c|}
\hline$x$ & $\lg x$ & $x$ & $\lg x$ \\
\hline 0.021 & $2 a+b+c-3$ & 6 & $1+a-b-c$ \\
\hline 0.27 & $6 a-3 b-2$ & 7 & $2(b+c)$ \\
\hline 1.5 & $3 a-b+c$ & 8 & $3-3 a-3 c$ \\
\hline 2.8 & $1-2 a+2 b-c$ & 9 & $4 a-2 b$ \\
\hline 3 & $2 a-b$ & 14 & $1-a+2 b$ \\
\hline 5 & $a+c$ & & \\
\hline
\end{tabular}
[Solution] The values of $\lg 1.5$ and $\lg 7$ are incorrect and should be corrected to $3 a-b+c-1$ and $2 b+c$, respectively.
The reasoning is as follows:
We start by analyzing the logarithms of the three prime numbers $3, 5, 7$:
(1) If $\lg 3 \neq 2 a-b$, then $\lg 9=2 \lg 3 \neq 4 a-2 b, \lg 0.27=3 \lg 3-2 \neq 6 a-3 b-2$, which means there are three incorrect logarithmic values in the table, contradicting the problem statement. Therefore,
$$
\lg 3=2 a-b
$$
(2) If $\lg 5 \neq a+c$, then $\lg 2=1-\lg 5 \neq 1-a-c$, and $\lg 8=3 \lg 2 \neq 3-3 a-3 c, \lg 6=\lg 3+\lg 2 \neq 1+a-b-c$, which also means there are three incorrect logarithmic values in the table, contradicting the problem statement. Therefore:
$$
\lg 5=a+c
$$

From (1) and (2), it is clear that the value of $\lg 1.5$ in the table is incorrect and should be corrected to:
$$
\lg 1.5=\lg 3+\lg 5-1=3 a-b+c-1 .
$$

Now, let's examine the values of $\lg 0.021, \lg 2.8, \lg 14$ in the table.
$$
\lg 0.021=\lg 3+\lg 7-3,
$$

Then
$$
\begin{array}{l}
2 a+b+c-3=(2 a-b)+(2 b+c)-3 ; \\
\lg 2.8=2 \lg 2+\lg 7-1, \\
1-2 a+2 b-c=2(1-a-c)+(2 b+c)-1 ; \\
\lg 14=\lg 2+\lg 7,
\end{array}
$$

Then

Then
$$
1-a+2 b=(1-a-c)+(2 b+c) \text {. }
$$

If $\lg 7=2(b+c)$, then the values of $\lg 0.021, \lg 2.8, \lg 14$ in the table are all incorrect, which again contradicts the problem statement. Therefore, the value of $\lg 7$ in the table is incorrect and should be corrected to:
$$
\lg 7=2 b+c .
$$