It's a very simple question and answer can be directly given 0 without lifting your pen.
$\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}.$$\begin{bmatrix} 1 & 9 & 5 \end{bmatrix}$
Column1 of Matrix A=$1 \times\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}$
Column2 of Matrix A=$9 \times \begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}$
Column3 of Matrix A=$5 \times \begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}$
All columns of A are linear combinations of $\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}$ and hence independent column is only column 1-->$\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}$
When an nxn matrix does not have full set of n independent columns
(1)Surely it's determinant is 0.
(2)One of the Eigen-values of such matrix should be 0.
(3)Such matrix won't be diagonalizable.
(4)On Reducing such matrix to Echelon form(U), while reducing, you will find a 0 entry at somewhere along the diagonal and hence this matrix will never be LU Decomposible.