Calculating variance for total data If I know the variance of column A, column B and Column C. But I want to know what is the variance of Column D which is equal to D = A+B+C. How do I calculate that variance? 
 A: I am suspicious about the meaning of the question.  Suppose you have two columns: $$ \left[ \begin{array}{c} 1 \\ 2 \\ 3 \\ 4 \\ 5 \end{array} \right], \left[ \begin{array}{c} 10 \\ 9 \\ 8 \\ 7 \\ 6 \end{array} \right] $$ Someone might observe that the average of the entries in the first column is $3$ and then find
$$
\left. \left(\begin{array}{r} (1-3)^2 \\ {} + (2-3)^2 \\ {} + (3-3)^2 \\ {} + (4-3)^2 \\ {} + (5-3)^2 \end{array} \right) \right/ 5, \text{ which comes to } 2.
$$
and call that the "variance" of the first column. Try that with both columns and and with their sum and you will find, in this particular case, that the variance of the sum is $0$ and the sum of the variances is $2+2=4 \ne0.$ The variance of the sum is not the sum of the variances, contrary to an assertion in a comment under the question. But, I imagine it will be said, these are not independent. Here's one thing that "independence" might mean in this case:
$$
\begin{array}{c|ccccc}
& 10 & 9 & 8 & 7 & 6 \\
\hline
1 \\
2 \\
3 \\
4 \\
5
\end{array}
$$
Fill in the addition table and get $25$ numbers. Then the variance of the "sum", e.g. the list of $25$ sums, is the sum of the two variances. But then would one call this a "sum of two columns" without explaining the meaning and expect it to be understood?
Another possibility is that the two columns are random variables and what is seen above is merely one realization of each of the two, and then the variance depends on their two separate probability distributions. Then one can speak of "independence". But in that case, what does "variance" mean? These are vector-valued random variables; what is their variance? The $5\times 5$ matrix of covariances of the entries? That's the standard thing, but I would expect the question to make clear that that is what is meant.
Another problem is that some people will say they were taught to divide by $5-1=4$ instead of $5$ where I divided by $5$ above. At best, that should be done only when estimating a population variance when what you have is a sample rather than the whole population. But if that is done, then the sum of the variances is not the variance of the sum.
So the question as it now stands is not very clear.
