Arithmetically progressive

Relevant wiki: Arithmetic Progressions

If $a , b , c$ is in arithmetic progression,
$-2a , -2b , -2c$ is in arithmetic progression,
So, $(a+b+c)-2a , (a+b+c)-2b , (a+b+c)-2c$ is in arithmetic progression,
So, $(b+c-a) , (c+a-b) , (a+b-c$ are in arithmetic progression.

So, the answer is $\color{#69047E}{\boxed{\text{True}}}$ .

Given that $a,b,c$ form an arithmetic progression, we know that

$d_1=c-b, \quad d_1=b-a$

Therefore, it satisfies $c-b=b-a$

Now, given a progression $(b+c-a),(c+a-b),(a+b-c)$

If it was an arithmetic progression:

$d_2 = (c+a-b) - (b+c-a) = 2a-2b, \quad d_2 = (a+b-c) - (c+a-b) = 2b-2c$

This gives $2a-2b = 2b-2c \implies -2(b-a) = -2(c-b) \implies b-a=c-b$

The second progression also satisfies the same condition, therefore we can say that it is an arithmetic progression too.

Therefore, the answer is $\boxed{\text{True}}$