The angle between the vector $\lambda i + 3j - 6k$ and the vector $i$ is $120^\circ$. Find the exact value of the constant $\lambda$.

1. Apply the scalar product formula to set up an equation involving $\lambda$.

Let $\mathbf{a} = \lambda\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}$ and $\mathbf{b} = \mathbf{i}$.
Using the scalar product formula $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$:
$(\lambda)(1) + (3)(0) + (-6)(0) = \sqrt{\lambda^2 + 3^2 + (-6)^2} \sqrt{1^2 + 0^2 + 0^2} \cos 120^\circ$
$\lambda = \sqrt{\lambda^2 + 45} (1) \left(-\frac{1}{2}\right)$

2. Rearrange and square both sides to eliminate the square root.

Multiply both sides by $-2$:
$-2\lambda = \sqrt{\lambda^2 + 45}$
Square both sides:
$(-2\lambda)^2 = (\sqrt{\lambda^2 + 45})^2$
$4\lambda^2 = \lambda^2 + 45$

3. Solve the resulting quadratic equation for $\lambda$.

$3\lambda^2 = 45$
$\lambda^2 = 15$
$\lambda = \sqrt{15} \text{ or } \lambda = -\sqrt{15}$

4. Reject the extraneous root using the context of the dot product.

Since the angle is obtuse ($120^\circ$), the scalar product $\mathbf{a} \cdot \mathbf{b} = \lambda$ must be negative.
Alternatively, from $-2\lambda = \sqrt{\lambda^2 + 45}$, the principal square root must be non-negative, which implies $\lambda \le 0$.
Therefore, we reject $\lambda = \sqrt{15}$.
$\lambda = -\sqrt{15}$