Let A=∣∣
∣
∣∣(a2+x2)abacab(b2+x2)bcacbc(c2+x2)∣∣
∣
∣∣
Using row operations, we get
R1→aR1,R2→bR2, R3→cR3⇒A=1abc∣∣
∣
∣∣a(a2+x2)a2ba2cab2b(b2+x2)b2cac2bc2c(c2+x2)∣∣
∣
∣∣⇒A=∣∣
∣
∣∣(a2+x2)a2a2b2(b2+x2)b2c2c2(c2+x2)∣∣
∣
∣∣
R1→R1+R2+R3⇒A=(a2+b2+c2+x2)∣∣
∣
∣∣111b2b2+x2b2c2c2(c2+x2)∣∣
∣
∣∣
C2→C2−C1, C3→C3−C1⇒A=(a2+b2+c2+x2)∣∣
∣
∣∣100b2x20c20x2∣∣
∣
∣∣⇒A=x4(a2+b2+c2+x2)
Hence, the maximum value of n=4