64

Let xyz be any number between 212 and 999

Let $x=2\to y+z=13$, then

(y, z) : (4, 9), (5, 8), (6, 7), (7, 6), (8, 5), (9, 4), i.e., 6 in number.

Let $x=3\to y+z=12$, then

(y, z) : (3, 9), (4, 8), (5, 7), (6, 6), (7, 5), (8, 4), i.e., 7 in number.

Let $x=4\to y+z=11$, then

(y, z) : (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4),(8, 3), (9, 2) i.e., 8 in number.

Let $x=5\to y+z=10$, then

(y, z) : (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3). (8, 2), (9, 1) i.e., 9 in number.

Let $x=6\to y+z=9$, then

(y, z) : (0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0) i.e., 10 in number.

Let $x=7\to y+z=8$, then

(y, z) : (0, 8), (1, 7), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (7, 1), (8, 0) i.e., 9 in number.

Let $x=8\to y+z=7$, then

(y, z) : (0, 7), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6,1), (7, 0) i.e., 8 in number.

Let $x=9\to y+z=6$, then

(y, z) : (0, 6), (1, 5), (2, 4), (3, 3), (4, 2), (5, 4), (6, 0) i.e., 7 in number.

Total = 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64.