To determine if each set $U$ of vectors is a basis for $\mathbb{R}^3$, we need to check two criteria for each set:

1. The vectors in $U$ must be linearly independent.
2. The vectors in $U$ must span $\mathbb{R}^3$.

These criteria are both satisfied if the set of vectors can be expressed as a matrix that has a determinant not equal to zero (or, equivalently, if the matrix is of full rank, which is 3 for $\mathbb{R}^3$).

Let's analyze each set $U$ given in the problem:

Set (a)

$U = \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix} \right\}$

Form the matrix and calculate the determinant: $A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 4 \end{bmatrix}$

Set (b)

$U = \left\{ \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \right\}$

Form the matrix and calculate the determinant: $B = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$

Set (c)

( U = \left{ \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}, \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}, \begin{bmatrix} 0 \ 0 \ 1 \