The problem asks us to find the value of the determinant given as $k$. The determinant is:

To find the value of $k$, we can analyze the properties of the determinant. A key property of determinants is that if two columns (or rows) are identical, the value of the determinant is zero.

Identifying Conditions for Zero Determinant

Let's examine the columns of the determinant:

• Column 1 ($C_1$): $\begin{pmatrix} \dfrac{1}{a(b+c)} \\ \dfrac{bc}{a(b+c)} \\ 1 \end{pmatrix}$
• Column 2 ($C_2$): $\begin{pmatrix} \dfrac{1}{b(c+a)} \\ \dfrac{ca}{b(c+a)} \\ 1 \end{pmatrix}$
• Column 3 ($C_3$): $\begin{pmatrix} \dfrac{1}{c(a+b)} \\ \dfrac{ab}{c(a+b)} \\ 1 \end{pmatrix}$

Consider the case where $a = b$. Let's substitute $b$ with $a$ in the determinant:

• The first column becomes: $\begin{pmatrix} \dfrac{1}{a(a+c)} \\ \dfrac{ac}{a(a+c)} \\ 1 \end{pmatrix}$
• The second column becomes: $\begin{pmatrix} \dfrac{1}{a(c+a)} \\ \dfrac{ca}{a(c+a)} \\ 1 \end{pmatrix}$

Comparing the first and second columns when $a=b$, we see that:

• The element in the first row of $C_1$ is $\dfrac{1}{a(a+c)}$.
• The element in the first row of $C_2$ is $\dfrac{1}{a(c+a)}$. These are identical.
• The element in the second row of $C_1$ is $\dfrac{ac}{a(a+c)}$.
• The element in the second row of $C_2$ is $\dfrac{ca}{a(c+a)}$. These are identical.
• The element in the third row of both columns is $1$.

Thus, when $a=b$, Column 1 ($C_1$) is identical to Column 2 ($C_2$).

Consequences of Identical Columns

According to the properties of determinants, if any two columns (or rows) of a determinant are identical, the value of the determinant is zero.

Therefore, when $a = b$, the determinant $k$ is $0$. Similarly:

• If $b = c$, Column 2 ($C_2$) becomes identical to Column 3 ($C_3$), making the determinant $k=0$.
• If $c = a$, Column 3 ($C_3$) becomes identical to Column 1 ($C_1$), making the determinant $k=0$.

Since the determinant evaluates to zero under these symmetric conditions ($a=b$, $b=c$, or $c=a$), it suggests that the determinant is identically zero for all valid values of $a, b,$ and $c$ (where denominators are non-zero).

Conclusion

Based on the analysis showing that the determinant becomes zero when $a=b$, $b=c$, or $c=a$, we conclude that the value of the determinant is always zero.

Therefore, $k=0$. This corresponds to option 4.