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The given line is:
$3x - y + 7 = 0 \implies y = 3x + 7$
Slope of this line, $m_1 = 3$.

Let the slope of the required line be $m$.
The angle between the two lines is $45^\circ$.
$\tan 45^\circ = \pm \frac{m - m_1}{1 + m \cdot m_1}$
$1 = \pm \frac{m - 3}{1 + 3m}$

Case 1: Taking the positive sign:
$1 = \frac{m - 3}{1 + 3m}$
$1 + 3m = m - 3$
$2m = -4 \implies m = -2$

Equation of the line passing through $(-1, 2)$ with slope $m = -2$:
$y - 2 = -2(x - (-1))$
$y - 2 = -2(x + 1)$
$2x + y = 0$

Case 2: Taking the negative sign:
$1 = -\frac{m - 3}{1 + 3m}$
$1 + 3m = 3 - m$
$4m = 2 \implies m = \frac{1}{2}$

Equation of the line passing through $(-1, 2)$ with slope $m = \frac{1}{2}$:
$y - 2 = \frac{1}{2}(x - (-1))$
$2(y - 2) = x + 1$
$2y - 4 = x + 1$
$x - 2y + 5 = 0$

Answer: $2x + y = 0$ or $x - 2y + 5 = 0$