Define two vectors forming two sides of the triangle from a common vertex.

Let the vertices be $P(1, 4, 6)$, $Q(-2, 5, -1)$, and $R(1, -1, 1)$.
We can form vectors $\vec{PQ}$ and $\vec{PR}$.

$$\vec{PQ} = Q - P = (-2-1, 5-4, -1-6) = (-3, 1, -7)$$ $$\vec{PR} = R - P = (1-1, -1-4, 1-6) = (0, -5, -5)$$

The area of a triangle with vertices $P, Q, R$ is given by $\frac{1}{2} |\vec{PQ} \times \vec{PR}|$.
First, calculate the cross product $\vec{PQ} \times \vec{PR}$:

$$\vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 1 & -7 \\ 0 & -5 & -5 \end{vmatrix}$$ $$= \mathbf{i}((1)(-5) - (-7)(-5)) - \mathbf{j}((-3)(-5) - (-7)(0)) + \mathbf{k}((-3)(-5) - (1)(0))$$ $$= \mathbf{i}(-5 - 35) - \mathbf{j}(15 - 0) + \mathbf{k}(15 - 0)$$ $$= -40\mathbf{i} - 15\mathbf{j} + 15\mathbf{k}$$

$$|\vec{PQ} \times \vec{PR}| = \sqrt{(-40)^2 + (-15)^2 + (15)^2}$$ $$= \sqrt{1600 + 225 + 225}$$ $$= \sqrt{2050}$$

We can simplify $\sqrt{2050}$:

$$\sqrt{2050} = \sqrt{25 \times 82} = 5\sqrt{82}$$

The area of the triangle is half the magnitude of the cross product:

$$\text{Area} = \frac{1}{2} |\vec{PQ} \times \vec{PR}| = \frac{1}{2} (5\sqrt{82})$$ $$\text{Area} = \frac{5\sqrt{82}}{2}$$

The area of the triangle is $\frac{5\sqrt{82}}{2}$.