If $$\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}, \vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$$ and $$\vec{c} = \lambda\hat{i} + \hat{j} + (2\lambda - 1)\hat{k}$$ are coplanar vectors, then $$\lambda$$ is equal to

The three vectors will be coplanar if and only if their scalar triple product is zero.

Definition: For $$\vec{a},\vec{b},\vec{c}$$ the scalar triple product is
$$\vec{a}\cdot(\vec{b}\times\vec{c})\;=\; \begin{vmatrix} a_x & a_y & a_z\\ b_x & b_y & b_z\\ c_x & c_y & c_z \end{vmatrix}.$$
Coplanarity condition: $$\vec{a}\cdot(\vec{b}\times\vec{c}) = 0.$$

Write each vector in component form:
$$\vec{a} = (1,\,-2,\,3),\; \vec{b} = (2,\,3,\,-1),\; \vec{c} = (\lambda,\,1,\,2\lambda-1).$$

Form the determinant and equate it to zero:
$$ \begin{vmatrix} 1 & -2 & 3\\ 2 & 3 & -1\\ \lambda & 1 & 2\lambda-1 \end{vmatrix} = 0.$$

Expand along the first row:

$$ 1\begin{vmatrix}3 & -1\\ 1 & 2\lambda-1\end{vmatrix} -(-2)\begin{vmatrix}2 & -1\\ \lambda & 2\lambda-1\end{vmatrix} +3\begin{vmatrix}2 & 3\\ \lambda & 1\end{vmatrix}=0. $$

Compute each 2×2 determinant:

1. $$\begin{vmatrix}3 & -1\\ 1 & 2\lambda-1\end{vmatrix} = 3(2\lambda-1) - (-1)(1) = 6\lambda - 3 + 1 = 6\lambda - 2.$$

2. $$\begin{vmatrix}2 & -1\\ \lambda & 2\lambda-1\end{vmatrix} = 2(2\lambda-1) - (-1)\lambda = 4\lambda - 2 + \lambda = 5\lambda - 2.$$ Multiplying by the prefactor gives $$-(-2) \times (\,5\lambda-2\,) = 2(5\lambda-2) = 10\lambda - 4.$$

3. $$\begin{vmatrix}2 & 3\\ \lambda & 1\end{vmatrix} = 2\cdot1 - 3\lambda = 2 - 3\lambda.$$ Multiplying by 3 gives $$3(2 - 3\lambda) = 6 - 9\lambda.$$

Add the three results and set the sum to zero:

$$(6\lambda - 2) + (10\lambda - 4) + (6 - 9\lambda) = 0$$
$$\Longrightarrow\; 7\lambda = 0.$$

Hence $$\lambda = 0.$$

Therefore the vectors are coplanar only when $$\lambda = 0.$$

Option A which is: 0