To solve for the three consecutive even integers that have a sum of 144, we can define the integers in terms of a variable.
Let the first even integer be \( x \). Then the next two consecutive even integers can be expressed as \( x + 2 \) and \( x + 4 \).
Now, we can set up the equation for their sum:
\[ x + (x + 2) + (x + 4) = 144 \]
Combining like terms, we get:
\[ 3x + 6 = 144 \]
Next, we can isolate \( x \) by subtracting 6 from both sides:
\[ 3x = 144 - 6 \] \[ 3x = 138 \]
Now, divide both sides by 3:
\[ x = \frac{138}{3} = 46 \]
So, the first even integer is \( 46 \). The other two consecutive even integers are:
\[ x + 2 = 46 + 2 = 48 \] \[ x + 4 = 46 + 4 = 50 \]
Thus, the three consecutive even integers are 46, 48, and 50.
The least of these integers is:
\[ \boxed{46} \]