To find the point in the span of $A = \text{Span}\{(2,0,1),(1,3,4)\}$ that is nearest to $\mathbf{b} = (5, 6, 7)$, we can use the concept of projection. Here's a step-by-step solution:

Step 1: Set Up the Problem

We need to find the vector $\mathbf{p}$ in the span of $A$ such that $\mathbf{p}$ is the closest point to $\mathbf{b}$.

Let the vectors in the span of $A$ be $\mathbf{v}_1 = (2, 0, 1)$ and $\mathbf{v}_2 = (1, 3, 4)$. Any vector in the span of $A$ can be written as: $\mathbf{p} = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 = c_1 (2, 0, 1) + c_2 (1, 3, 4)$ $\mathbf{p} = (2c_1 + c_2, 3c_2, c_1 + 4c_2)$

Step 2: Formulate the Problem as a System of Equations

We want to minimize the distance between $\mathbf{p}$ and $\mathbf{b}$. The distance (squared) between $\mathbf{p}$ and $\mathbf{b}$ is: $\|\mathbf{p} - \mathbf{b}\|^2 = (2c_1 + c_2 - 5)^2 + (3c_2 - 6)^2 + (c_1 + 4c_2 - 7)^2$

Step 3: Solve for $c_1$ and $c_2$

To find the minimum, we take the partial derivatives of $\|\mathbf{p} - \mathbf{b}\|^2$ with respect to $c_1$ and $c_2$, and set them to zero.

$\frac{\partial}{\partial c_1} \left[ (2c_1 + c_2 - 5)^2 + (3c_2 - 6)^2 + (c_1 + 4c_2 - 7)^2 \right] = 0$ $\frac{\partial}{\partial c_2} \left[ (2c_1 + c_2 - 5)^2 + (3c_2 - 6)^2 + (c_1 + 4c_2 - 7)^2 \right] = 0$

This results in the following system of linear equations:

1. $2(2c_1 + c_2 - 5) \cdot 2 + 2(c_1 + 4c_2 - 7) \cdot 1 = 0$
2. $2(2c_1 + c_2 - 5) \cdot 1 + 2(3c_2 - 6) \cdot 3 + 2(c_1 + 4c_2 - 7) \cdot 4 = 0$

Step 4: Simplify the Equations

Simplify the equations:

1. $8c_1 + 4c_2 - 20 + 2c_1 + 8c_2 - 14 = 0$ $10c_1 + 12c_2 = 34$ $5c_1 + 6c_2 = 17 \quad \text{(Equation 1)}$

2. $4c_1 + 2c_2 - 10 + 18c_2 - 36 + 4c_1 + 16c_2 - 28 = 0$ $8c_1 + 36c_2 = 74$ $4c_1 + 18c_2 = 37 \quad \text{(Equation 2)}$

Step 5: Solve the System of Equations

Solve the linear system: $5c_1 + 6c_2 = 17$ $4c_1 + 18c_2 = 37$

From Equation 1: [ c_1 = \frac{