If $$a, b, c$$ are non zero complex numbers satisfying $$a^2 + b^2 + c^2 = 0$$ and $$$\begin{vmatrix} b^2+c^2 & ab & ac \\ ab & c^2+a^2 & bc \\ ac & bc & a^2+b^2 \end{vmatrix} = ka^2 b^2 c^2$$$, then $$k$$ is equal to

Given $$a^2+b^2+c^2=0.$$

Therefore, $$b^2+c^2=-a^2,\quad c^2+a^2=-b^2,\quad a^2+b^2=-c^2.$$

Hence the determinant becomes $$D=\begin{vmatrix} -a^2 & ab & ac \\ ab & -b^2 & bc \\ ac & bc & -c^2 \end{vmatrix}.$$

Taking $$abc$$ common from each row, we get $$D=a^2b^2c^2\begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix}.$$

Let $$\Delta=\begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix}.$$

Applying the row operations $$R_2\to R_2+R_1$$ and $$R_3\to R_3+R_1,$$ we obtain $$\Delta=\begin{vmatrix} -1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \end{vmatrix}.$$

Expanding along the first column, $$\Delta=(-1)\begin{vmatrix} 0 & 2 \\ 2 & 0 \end{vmatrix}.$$

Therefore, $$\Delta=(-1)(0-4)=4.$$

Hence, $$D=4a^2b^2c^2.$$

Comparing with $$D=ka^2b^2c^2,$$ we get $$k=4.$$