向量 $\mathbf{a}$、$\mathbf{b}$ 和 $\mathbf{c}$ 满足 $\mathbf{b} \ne 0$ 且 $3\mathbf{a} \times \mathbf{b} = 7\mathbf{b} \times \mathbf{c}$。

1. Use the anticommutative property of the vector product to rewrite the right side of the equation.

Given $3\mathbf{a} \times \mathbf{b} = 7\mathbf{b} \times \mathbf{c}$.
Since $\mathbf{b} \times \mathbf{c} = -\mathbf{c} \times \mathbf{b}$, we have:
$3\mathbf{a} \times \mathbf{b} = -7\mathbf{c} \times \mathbf{b}$

2. Rearrange the terms and factorise using the distributive property.

$3\mathbf{a} \times \mathbf{b} + 7\mathbf{c} \times \mathbf{b} = \mathbf{0}$
$(3\mathbf{a} + 7\mathbf{c}) \times \mathbf{b} = \mathbf{0}$

3. Deduce the relationship between the vectors from the zero vector product.

The vector product of $(3\mathbf{a} + 7\mathbf{c})$ and $\mathbf{b}$ is the zero vector. Since it is given that $\mathbf{b} \ne \mathbf{0}$, it must be that the vector $(3\mathbf{a} + 7\mathbf{c})$ is parallel to $\mathbf{b}$ (or is the zero vector itself).
Therefore, we can write:
$3\mathbf{a} + 7\mathbf{c} = \lambda \mathbf{b}$
where $\lambda$ is a scalar constant. (Shown)

1. Take the scalar product of the vector equation from part (i) with itself.

Using the result from (i), $3\mathbf{a} + 7\mathbf{c} = \lambda \mathbf{b}$. Taking the scalar product of both sides with themselves:
$(3\mathbf{a} + 7\mathbf{c}) \cdot (3\mathbf{a} + 7\mathbf{c}) = (\lambda \mathbf{b}) \cdot (\lambda \mathbf{b})$
$9|\mathbf{a}|^2 + 42(\mathbf{a} \cdot \mathbf{c}) + 49|\mathbf{c}|^2 = \lambda^2 |\mathbf{b}|^2$

2. Evaluate the scalar product $\mathbf{a} \cdot \mathbf{c}$.

Using the definition of the scalar product with the given angle and magnitudes:
$\mathbf{a} \cdot \mathbf{c} = |\mathbf{a}| |\mathbf{c}| \cos 60^\circ$
$\mathbf{a} \cdot \mathbf{c} = (1)\left(\frac{3}{7}\right)\left(\frac{1}{2}\right) = \frac{3}{14}$

3. Substitute the known values into the expanded equation to solve for $\lambda$.

Substitute $|\mathbf{a}| = 1$, $|\mathbf{b}| = 1$, $|\mathbf{c}| = \frac{3}{7}$ and $\mathbf{a} \cdot \mathbf{c} = \frac{3}{14}$ into the expanded equation:
$9(1)^2 + 42\left(\frac{3}{14}\right) + 49\left(\frac{3}{7}\right)^2 = \lambda^2 (1)^2$
$9 + 3(3) + 49\left(\frac{9}{49}\right) = \lambda^2$
$9 + 9 + 9 = \lambda^2$
$\lambda^2 = 27$
$\lambda = \pm\sqrt{27} = \pm 3\sqrt{3}$