Find the constant $k$ from the determinant 
Given:  $$\begin{vmatrix}(b+c)^2 &a^2&a^2\\b^2 &(c+a)^2&b^2 \\c^2&c^2& (a+b)^2\end{vmatrix}=k(abc)(a+b+c)^3$$ Find $k$.

If I directly open the determinant it will go to long I can't apply most of the row or column operation as they keep making it more complex.

It is also simple to compute the posted determinant, $D$ for short, thus ignoring the information on the special form of the result. We are subtracting at the first step the first column from the other two.
$$
\begin{aligned}
D &=
\begin{vmatrix}
(b+c)^2 & a^2 & a^2 \\
b^2 & (c+a)^2 & b^2 \\
c^2 & c^2 & (a+b)^2
\end{vmatrix}
\\[2mm]
&=
\begin{vmatrix}
(b+c)^2 & a^2- (b+c)^2 & a^2 - (b+c)^2 \\
b^2 & (c+a)^2-b^2 & 0 \\
c^2 & 0 & (a+b)^2-c^2
\end{vmatrix}
\\[2mm]
&=
\begin{vmatrix}
(b+c)^2 & a- b-c & a- b-c \\
b^2 & c+a-b & 0 \\
c^2 & 0 & a+b-c
\end{vmatrix}
\cdot(a+b+c)^2
\\[2mm]
&=
\begin{vmatrix}
2bc & -2c & - 2b \\
b^2 & c+a-b & 0 \\
c^2 & 0 & a+b-c
\end{vmatrix}
\cdot(a+b+c)^2
\\[2mm]
&=
\begin{vmatrix}
2bc-bc-bc & -2c & - 2b \\
b^2+\frac b2(c+a-b)+\frac c2\cdot 0 & c+a-b & 0 \\
c^2+\frac c2\cdot 0+\frac c2(a+b-c) & 0 & a+b-c
\end{vmatrix}
\cdot(a+b+c)^2
\\[2mm]
&=
\begin{vmatrix}
0 & -2c & - 2b \\
\frac b2 & c+a-b & 0 \\
\frac c2 & 0 & a+b-c
\end{vmatrix}
\cdot(a+b+c)^3
\\[2mm]
&=
-\begin{vmatrix}
0 & c & b \\
b & c+a-b & 0 \\
c & 0 & a+b-c
\end{vmatrix}
\cdot(a+b+c)^3
\\[2mm]
&=2abc(a+b+c)^3\ .
\end{aligned}
$$
(At the last step, using Sarrus, only two out of six terms survive, we can factor $bc$, and $-(a+b-c)-(c+a-b)=-2a$ leads immediately to the result. No step was omitted, symmetry was used when possible.)

Computer check:
sage: var('a,b,c');
sage: A = matrix(3,3, [(b+c)^2, a^2, a^2, b^2, (a+c)^2, b^2, c^2, c^2, (a+b)^2] )
sage: A.det().factor()
2*(a + b + c)^3*a*b*c

