Determine whether the vectors v₁ = (1, 0), v₂ = (0, 1) are linearly independent in R².

Expanding the vector equation gives:

$$(3, 7, -4) = (a_1 + 2a_2 + 3a_3, 2a_1 + 3a_2 + 5a_3, 3a_1 + 7a_2 + 6a_3)$$

Equating the components, we get the following system:
\begin{align*} a_1 + 2a_2 + 3a_3 &= 3 \ 2a_1 + 3a_2 + 5a_3 &= 7 \ 3a_1 + 7a_2 + 6a_3 &= -4 \end{align*}

Apply $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$:

$$\begin{bmatrix} 1 & 2 & 3 & | & 3 \\ 0 & -1 & -1 & | & 1 \\ 0 & 1 & -3 & | & -13 \end{bmatrix}$$

Apply $R_3 \to R_3 + R_2$:

$$\begin{bmatrix} 1 & 2 & 3 & | & 3 \\ 0 & -1 & -1 & | & 1 \\ 0 & 0 & -4 & | & -12 \end{bmatrix}$$

From the third row:

$$-4a_3 = -12 \implies a_3 = \frac{-12}{-4} \implies a_3 = 3$$

From the second row:

$$-a_2 - a_3 = 1$$ $$-a_2 - 3 = 1 \implies -a_2 = 4 \implies a_2 = -4$$

From the first row:

$$a_1 + 2a_2 + 3a_3 = 3$$ $$a_1 + 2(-4) + 3(3) = 3$$ $$a_1 - 8 + 9 = 3$$ $$a_1 + 1 = 3 \implies a_1 = 2$$

The scalars are $a_1 = 2$, $a_2 = -4$, and $a_3 = 3$.

The final answer is $a_1 = 2, a_2 = -4, a_3 = 3$.