If a, b, c, are non zero complex numbers satisfying a^2+b^2+c^2=0 and | … |=k a^2 b^2 c^2, then k is equal to

Let $\Delta=\left|\begin{array}{ccc}b^2+c^2 & a b & a c \\ a b & c^2+a^2 & b c \\ a c & b c & a^2+b^2\end{array}\right|$Δ=​b2+c2abac​abc2+a2bc​acbca2+b2​​ Multiply $C_1$C1​ by $a, C_2$a,C2​ by b and $C_3$C3​ by $\mathrm{c}$c and hence divide by $a b c$abc. $$=\frac{1}{a b c}\left|\begin{array}{ccc} a\left(b^2+c^2\right. & \left.a b^2\right) & a c^2 \\ a^2 b & b\left(c^2+a^2\right. & \left.b c^2\right) \\ a^2 c & b^2 c & c\left(a^2+b^2\right. \end{array}\right|$$=abc1​​a(b2+c2a2ba2c​ab2)b(c2+a2b2c​ac2bc2)c(a2+b2​​ Take out $a, b, c$a,b,c common from $R_1, R_2$R1​,R2​ and $R_3$R3​ respectively. $$\begin{aligned} & \therefore \quad \Delta=\frac{a b c}{a b c}\left|\begin{array}{ccc} b^2+c^2 & b^2 & c^2 \\ a^2 & c^2+a^2 & c^2 \\ a^2 & b^2 & a^2+b^2 \end{array}\right| \\ & \text { Apply } C_1 \rightarrow C_1-C_2-C_3 \\ & \Delta=\left|\begin{array}{ccc} 0 & b^2 & c^2 \\ -2 c^2 & c^2+a^2 & c^2 \\ -2 b^2 & b^2 & a^2+b^2 \end{array}\right| \\ & =-2\left|\begin{array}{ccc} 0 & b^2 & c^2 \\ c^2 & c^2+a^2 & c^2 \\ b^2 & b^2 & a^2+b^2 \end{array}\right| \\ & \end{aligned}$$​∴Δ=abcabc​​b2+c2a2a2​b2c2+a2b2​c2c2a2+b2​​ Apply C1​→C1​−C2​−C3​Δ=​0−2c2−2b2​b2c2+a2b2​c2c2a2+b2​​=−2​0c2b2​b2c2+a2b2​c2c2a2+b2​​​ Apply $C_2-C_1$C2​−C1​ and $C_3-C_1$C3​−C1​ $$\begin{aligned} & =-2\left|\begin{array}{ccc} 0 & b^2 & c^2 \\ c^2 & a^2 & 0 \\ b^2 & 0 & a^2 \end{array}\right| \\ & =-2\left[-b^2\left(c^2 a^2\right)+c^2\left(-a^2 b^2\right)\right] \\ & =2 a^2 b^2 c^2+2 a^2 b^2 c^2=4 a^2 b^2 c^2 \\ & \text { But } \Delta=k a^2 b^2 c^2 \therefore k=4 \end{aligned}$$​=−2​0c2b2​b2a20​c20a2​​=−2[−b2(c2a2)+c2(−a2b2)]=2a2b2c2+2a2b2c2=4a2b2c2 But Δ=ka2b2c2∴k=4​