When a, B and C are integers and satisfy A2 + B2 + C2 + 4a-8b-14c + 69 = 0, the value of a + 2b-3c is obtained

When a, B and C are integers and satisfy A2 + B2 + C2 + 4a-8b-14c + 69 = 0, the value of a + 2b-3c is obtained

a2＋b2＋c2＋4a－8b－14c＋69＝0
（a²+4a+4）+（b²-8b+16）+（c²-14c+49）=0
（a+2）²+（b-4）²+（c-7）²=0
a= -2,b=4,c=7
a+2b-3c= -2+2x4-3x7= -2+8-21= -15

If a + 2B + C = 0, 2A + 5B + 3C = 0 (where B ≠ 0), then the value of AB + BC + AC / A2 + B2 + C2 is

It seems that the problem is a little vague~

Let the opposite sides of the internal angles a, B and C of the triangle ABC be a, B and C respectively. If a = π / 3 and a = root 3, then the value range of B ^ 2 + C ^ 2 is obtained
Above: the answer is (3,6]

b²+c²-2bccosa=a²
b²+c²-bc=3
Because B & # 178; + C & # 178; > = 2BC
So B & # 178; + C & # 178; - (B & # 178; + C & # 178;) / 2BC + 3 > 3
So B & # 178; + C & # 178; ∈ (3,6]