$\begin{array}{1 1} 55 \\ 72 \\ 110 \\ 60 \end{array}$

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Combination of 3 negative integers whose sum is -12 are:

$(-1,-1,-10),(-1,-2,-9),(-1,-3,-8),(-1,-4,-7),(-1,-5,-6)$

$(-2,-2,-8),(-2,-3,-7),(-2,-4,-6),(-2,-5,-5)$

$(-3,-3,-6),(-3,-4,-5)$

$(-4,-4,-4)$

No. of points with the combination $(-4,-4,-4) = 1$

No. of points with the combinations $ (-1,-1,-10),(-2,-2,-8),(-2,-5,-5),(-3,-3,-6)$

= 3 in each combination=$3\times 4=12$

Remaining 7 combinations have $3!=6$ points in each combination.

$=7\times 6=42$ points

$\therefore$ Total number of required points in space =

$1+12+42=55$

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