Find the real number $k$ such that: $\det\begin{bmatrix} 2a_{1} & 2a_{2} & 2a_{3}\\ 3b_{1}+5c_{1} & 3b_{2}+5c_{2} & 3b_{3}+5c_{3}\\ 7c_{1} & 7c_{2} & 7c_{3} \end{bmatrix} = k * \det\begin{bmatrix} a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3} \end{bmatrix}$
knowing that $\det\begin{bmatrix} a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3} \end{bmatrix} ≠ 0$
I have calculated the table in terms of the first/second and third column. Then I wanted to use the Sarrus rule for the other matrix, but I am stuck somewhere there.
Check out this helpful link which tells you how elementary row operations affect the determinant.
Applying $R_2 \rightarrow R_2 - \dfrac57 R_3$, the determinant is unchanged, and we get:
$$\left| \begin{matrix} 2a_1 & 2a_2 & 2a_3 \\ 3b_1 & 3b_2 &3b_3 \\ 7c_1 & 7c_2 & 7c_3 \end{matrix} \right|$$
Now applying $R_1 \rightarrow \dfrac{1}{2} R_1$, $R_2 \rightarrow \dfrac{1}{3} R_2$, and $R_3 \rightarrow \dfrac{1}{7} R_3$ render the determinant as follows:
$$2\times 3 \times 7 \times \left| \begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 &b_3 \\ c_1 & c_2 & c_3 \end{matrix} \right|$$
So the $k$ you're looking for is $2\times 3 \times 7 = 42$.
Edit: note that if you have a matrix $A$ and you multiply one of the rows of $A$ by $\lambda$ to get a new matrix $B$, then you have $|B| = \lambda |A|$. Hence, if I start with a $3\times 3$ matrix $A$ as in this question and I apply successive multiplications of its rows by $\lambda_1$, $\lambda_2$ and $\lambda_3$ to get a new matrix $B$, what we would have is:
$$|B| = \lambda_1 \lambda_2 \lambda_3 |A|$$
so that
$$|A| = \dfrac{1}{\lambda_1 \lambda_2 \lambda_3}|B|$$
which justifies why it's $2 \times 3 \times 7$ instead of $\dfrac{1}{2\times 3 \times 7}$.