If $a^{-1}+b^{-1}+c^{-1}=0$ and $\begin{vmatrix}1+a&1&1\ 1&1+b&1\ 1&1&1+c\end{vmatrix}=k$, then :

Question

Accepted Answer

To solve the problem, we need to evaluate the determinant of the given matrix and verify why Option 3 is the correct answer.

The matrix given is:
$$
\begin{vmatrix}
1+a & 1 & 1 \
1 & 1+b & 1 \
1 & 1 & 1+c
\end{vmatrix}
$$

The determinant of a 3x3 matrix $\begin{vmatrix} p & q & r \ s & t & u \ v & w & x \end{vmatrix}$ is calculated as:
$$ p(t \cdot x - u \cdot w) - q(s \cdot x - u \cdot v) + r(s \cdot w - t \cdot v) $$

Applying this formula to our matrix:
$$
\begin{vmatrix}
1+a & 1 & 1 \
1 & 1+b & 1 \
1 & 1 & 1+c
\end{vmatrix}
$$

The determinant is:
$$
(1+a)((1+b)(1+c) - 1 \cdot 1) - 1(1 \cdot (1+c) - 1 \cdot 1) + 1(1 \cdot 1 - (1+b) \cdot 1)
$$

Simplifying each term:
1. $(1+b)(1+c) - 1 = 1 + b + c + bc - 1 = b + c + bc$
2. $1 \cdot (1+c) - 1 = c$
3. $1 - (1+b) = -b$

Substituting back, we get:
$$
(1+a)(b + c + bc) - c - b
$$

Expanding $(1+a)(b + c + bc)$:
$$
= (b + c + bc) + a(b + c + bc) = b + c + bc + ab + ac + abc
$$

Thus, the determinant becomes:
$$
b + c + bc + ab + ac + abc - c - b
$$

Simplifying:
$$
bc + ab + ac + abc
$$

Given that $a^{-1} + b^{-1} + c^{-1} = 0$, we have:
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \implies \frac{ab + bc + ca}{abc} = 0 \implies ab + bc + ca = 0
$$

Substituting $ab + bc + ca = 0$ into the determinant expression:
$$
bc + ab + ac + abc = abc
$$

Thus, the determinant simplifies to $abc$, which confirms that $k = abc$.

**Explanation for Other Options:**

1. **Option 1: $k = (a+b+c)$** - This is incorrect because the determinant simplifies to $abc$, not $a+b+c$.

2. **Option 2: $k = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$** - This is incorrect because $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$, and the determinant simplifies to $abc$.

3. **Option 4: $k = 0$** - This is incorrect because the determinant simplifies to $abc$, not zero.

Therefore, the correct answer is Option 3: $k = abc$.