painting a house

Andy can finish the work in $A$ days. Ben can finish the work in $B$ days. Chris can finish the work in $C$ days.

In one day, Andy can finish $\dfrac{1}{A}$ work. Similar for Ben and Chris. Given,

$\begin{cases} \dfrac{1}{A} + \dfrac{1}{B} & = \dfrac{1}{10} \space ...(1) \\ \dfrac{1}{A} + \dfrac{1}{C} & = \dfrac{1}{12} \space ...(2) \\ \dfrac{1}{B} + \dfrac{1}{C} & = \dfrac{1}{20} \space ...(3) \end{cases}$

In one day, Andy and Ben can finish $\dfrac{1}{10}$ work, together. And similar for Ben & Chris, and Andy & Chris.

$(1) + (2) + (3)$ gives

$\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C} = \dfrac{7}{60} \space ...(4)$ .

$(4) - (1)$ gives

$\dfrac{1}{C} = \dfrac{1}{60}$

Hence, the answer is $C = 60$ .

Let the work be $W$ and the rates of doing work of $A$ , $B$ and $C$ be $a$ , $b$ and $c$ respectively. Then we have:

$\begin{cases} 10(a+b) = W & \implies a+b = \dfrac W{10} & ...(1) \\ 12(c+a) = W & \implies c+a = \dfrac W{12} & ...(2) \\ 10(a+b) = W & \implies b+c = \dfrac W{20} & ...(3) \end{cases}$

$\begin{aligned} (3)+(2)-(1): \quad 2c & = \frac W{20} + \frac W{12} - \frac W{10} \\ & = W \left(\frac {3+5-6}{60}\right) \\ 60c & = W \end{aligned}$

It will take $C$ $\boxed{60}$ days to do the work alone.

Let's do it without Algebra. Let Andy be A, Ben be B and Chris be C.
Let's suppose total area to be painted is 60 units. (LCM of 10, 12 and 20)
A+B can paint 6 units per day
A+C can paint 5 units per day
B+C can paint 3 units per day
From first and second statement we see that B can paint one more unit than C in one day. When we look at the third statement it is obvious that C paints 1 unit per day, B paints 2 units per day. It can be further confirmed that A paints 4 units per day though that is not needed for our solution. So C will need 60 days.