Here is Prob. 2.43 in the book Vector Analysis & Cartesian Tensors by D. E. Bourne and P. C. Kendall, 3rd edition:
If $$ \begin{align} \mathbf{A}_1 &= \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} + \nu_1 \mathbf{c}, \\ \mathbf{A}_2 &= \lambda_2 \mathbf{a} + \mu_2 \mathbf{b} + \nu_2 \mathbf{c}, \\ \mathbf{A}_3 &= \lambda_3 \mathbf{a} + \mu_3 \mathbf{b} + \nu_3 \mathbf{c}, \end{align} $$ show that $$ \mathbf{A}_1 \cdot \left( \mathbf{A}_2 \times \mathbf{A}_3 \right) = \left| \begin{matrix} \lambda_1 & \mu_1 & \nu_1 \\ \lambda_2 & \mu_2 & \nu_2 \\ \lambda_3 & \mu_3 & \nu_3 \end{matrix} \right| \mathbf{a} \cdot ( \mathbf{b} \times \mathbf{c} ). $$
My Attempt:
In what follows, we will be using the following identities: $$ \mathbf{a} \cdot ( \mathbf{a} \times \mathbf{b} ) = \mathbf{b} \cdot ( \mathbf{a} \times \mathbf{b} ) = \mathbf{b} \cdot ( \mathbf{b} \times \mathbf{c} ) = \mathbf{c} \cdot ( \mathbf{b} \times \mathbf{c} ) = \mathbf{c} \cdot ( \mathbf{c} \times \mathbf{a} ) = \mathbf{a} \cdot ( \mathbf{c} \times \mathbf{a} ) = 0, $$ and also that $$ \mathbf{a} \cdot ( \mathbf{b} \times \mathbf{c} ) = \mathbf{b} \cdot ( \mathbf{c} \times \mathbf{a} ) = \mathbf{c} \cdot ( \mathbf{a} \times \mathbf{b} ). $$
Using the distributive property of the dot and the cross products, we obtain $$ \begin{align} & \qquad \mathbf{A}_1 \cdot \left( \mathbf{A}_2 \times \mathbf{A}_3 \right) \\ &= \left( \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} + \nu_1 \mathbf{c} \right) \cdot \left[ \left( \lambda_2 \mathbf{a} + \mu_2 \mathbf{b} + \nu_2 \mathbf{c} \right) \times \left( \lambda_3 \mathbf{a} + \mu_3 \mathbf{b} + \nu_3 \mathbf{c} \right) \right] \\ &= \left( \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} + \nu_1 \mathbf{c} \right) \cdot \left[ \left( \lambda_2 \mu_3 - \lambda_3 \mu_2 \right) ( \mathbf{a} \times \mathbf{b} ) + \left( \mu_2 \nu_3 - \mu_3 \nu_2 \right) ( \mathbf{b} \times \mathbf{c} ) + \left( \nu_2 \lambda_3 - \nu_3 \lambda_2 \right) ( \mathbf{c} \times \mathbf{a} ) \right] \\ &= \lambda_1 \left( \lambda_2 \mu_3 - \lambda_3 \mu_2 \right) \big[ \mathbf{a} \cdot ( \mathbf{a} \times \mathbf{b} ) \big] + \lambda_1 \left( \mu_2 \nu_3 - \mu_3 \nu_2 \right) \big[ \mathbf{a} \cdot ( \mathbf{b} \times \mathbf{c} ) \big] + \lambda_1 \left( \nu_2 \lambda_3 - \nu_3 \lambda_2 \right) \big[ \mathbf{a} \cdot ( \mathbf{c} \times \mathbf{a} ) \big] \\ & \qquad + \mu_1 \left( \lambda_2 \mu_3 - \lambda_3 \mu_2 \right) \big[ \mathbf{b} \cdot ( \mathbf{a} \times \mathbf{b} ) \big] + \mu_1 \left( \mu_2 \nu_3 - \mu_3 \nu_2 \right) \big[ \mathbf{b} \cdot ( \mathbf{b} \times \mathbf{c} ) \big] + \mu_1 \left( \nu_2 \lambda_3 - \nu_3 \lambda_2 \right) \big[ \mathbf{b} \cdot ( \mathbf{c} \times \mathbf{a} ) \big] \\ & \qquad + \nu_1 \left( \lambda_2 \mu_3 - \lambda_3 \mu_2 \right) \big[ \mathbf{c} \cdot ( \mathbf{a} \times \mathbf{b} ) \big] + \nu_1 \left( \mu_2 \nu_3 - \mu_3 \nu_2 \right) \big[ \mathbf{c} \cdot ( \mathbf{b} \times \mathbf{c} ) \big] + \nu_1 \left( \nu_2 \lambda_3 - \nu_3 \lambda_2 \right) \big[ \mathbf{c} \cdot ( \mathbf{c} \times \mathbf{a} ) \big] \\ &= \lambda_1 \left( \mu_2 \nu_3 - \mu_3 \nu_2 \right) \big[ \mathbf{a} \cdot ( \mathbf{b} \times \mathbf{c} ) \big] + \mu_1 \left( \nu_2 \lambda_3 - \nu_3 \lambda_2 \right) \big[ \mathbf{b} \cdot ( \mathbf{c} \times \mathbf{a} ) \big] + \nu_1 \left( \lambda_2 \mu_3 - \lambda_3 \mu_2 \right) \big[ \mathbf{c} \cdot ( \mathbf{a} \times \mathbf{b} ) \big] \\ &= \left[ \lambda_1 \left( \mu_2 \nu_3 - \mu_3 \nu_2 \right) + \mu_1 \left( \nu_2 \lambda_3 - \nu_3 \lambda_2 \right) + \nu_1 \left( \lambda_2 \mu_3 - \lambda_3 \mu_2 \right) \right] \big[ \mathbf{a} \cdot ( \mathbf{b} \times \mathbf{c} ) \big] \\ &= \left| \begin{matrix} \lambda_1 & \mu_1 & \nu_1 \\ \lambda_2 & \mu_2 & \nu_2 \\ \lambda_3 & \mu_3 & \nu_3 \end{matrix} \right| \mathbf{a} \cdot ( \mathbf{b} \times \mathbf{c} ), \end{align} $$ as required.
Is this solution correct? If so, is it clear enough? Or, do I need to include some more details?
Your approach is correct, but there is a simpler proof using triple product formula:
$$\vec{u}.(\vec{v} \times \vec{w})=\det(u,v,w)\tag{1}$$
making the relationship to be proven equivalent to:
$$\det(A_1,A_2,A_3) \overset{?}{=} \begin{vmatrix} \lambda_1 & \mu_1 & \nu_1 \\ \lambda_2 & \mu_2 & \nu_2 \\ \lambda_3 & \mu_3 & \nu_3 \end{vmatrix} \det(a,b,c)\tag{2}$$
Now, your initial data with 3 relationships can be grouped into a single matrix relationship:
$$\underbrace{(A_1|A_2|A_3)}_{3 \times 3 \ \text{matrix}}=\underbrace{(a|b|c)}_{3 \times 3 \ \text{matrix}}\left(\begin{matrix} \lambda_1 & \lambda_2 & \lambda_3 \\ \mu_1 & \mu_2 & \mu_3 \\ \nu_1 & \nu_2 & \nu_3 \end{matrix} \right) \tag{3}$$
How to prove that (3) implies (2) ? Plainly by taking the determinant of both sides... (please note that in (3) we had to take the transpose of the matrix of coefficients, which is harmless for the computation of its determinant).