ANSWER
The determinant is 0
EXPLANATION (METHOD 1)
This method involves expanding along any column.
For an n×n matrix A, the determinant of A, det(A), can be obtained by expanding along the kth column:

where
is the entry of A in the kth row, 1st column,
is the entry of A in the kth row, 2nd column, etc., and
is the kn cofactor of A, defined as
.
But we do not need to care about the cofactors as all the 2nd column entries are
a₁₂ = a₂₂ = a₃₂ = 0
We would end up with

EXPLANATION (METHOD 2)|
This method involves expanding along a row
For an n×n matrix A, the determinant of A, det(A), can be obtained by expanding along the kth row:

where
is the entry of A in the kth row, 1st column,
is the entry of A in the kth row, 2nd column, etc., and
is the kn cofactor of A, defined as
.
is the kn minor, obtained by getting the determinant of the matrix which is the matrix A with row k and column n deleted.
Applying this here, we can expand along the 1st row.
For convenience, let G be the coefficient matrix of this question, which is

with the first row bolded.
The determinant is therefore

Note that g₁₁ is the matrix element of G that is in the 1st row, 1st column, g₁₂ is the matrix element of G that is in the 1st row, 2nd column, etc. Then we have

M₁₁ is the determinant of the matrix that is matrix G with row 1 and column 1 removed. The bold entires are the row and the column we delete.

The determinant of a 2×2 matrix is

so it follows that

Applying the same for M₁₂ and M₁₃, we have

and

so therefore
