Answer
See below
Work Step by Step
Let $A = [a_1, a_2,..., a_n]$ be an $n \times n$ matrix, and let $b = c_1a_1 + c_2a_2 +···+ c_na_n$, where $c_1, c_2,..., c_n$ are constants.
Obtain: $B_k=[a_1, a_2,..., a_{k-1},c_1a_1+c_2a_2+...c_na_n,a_{k+1},...,a_n]\\
\rightarrow \det(B_k)=([a_1, a_2,..., a_{k-1},c_1a_1+c_2a_2+...c_na_n,a_{k+1},...,a_n])$
Using property P6:
$\det(B_k)=\det[a_1, a_2,..., a_{k-1},c_1a_1,a_{k+1},...,a_n]+\det[a_1, a_2,..., a_{k-1},c_2a_2,a_{k+1},...,a_n]+...\det [a_1, a_2,..., a_{k-1},c_na_n,a_{k+1},...,a_n]$
Using property P2:
$\det(B_k)=c_1 \det[a_1, a_2,..., a_{k-1},a_1,a_{k+1},...,a_n]+c_2\det[a_1, a_2,..., a_2,a_{k+1},...,a_n]+...c_n\det [a_1, a_2,..., a_{k-1},a_n,a_{k+1},...,a_n]$
All of the above determinants are all zeroes, according to property P7.
Hence, $\det(B_k)\\=c_k\det [a_1, a_2,..., a_{k-1},a_k,a_{k+1},...,a_n]\\
=c_k\det(A)$
