Find the slope of the given line.

The equation of the given line is $3y = 5 - 2x$.
To find its slope, rewrite it in the slope-intercept form $y = mx + c$:

$$3y = -2x + 5$$ $$y = -\frac{2}{3}x + \frac{5}{3}$$

The slope of this line, let's call it $m_1$, is $-\frac{2}{3}$.

The line passing through points A $(2, 1)$ and B $(4, m)$ is perpendicular to the given line.
If two lines are perpendicular, the product of their slopes is $-1$.
Let $m_2$ be the slope of the line through A and B.

$$m_1 \times m_2 = -1$$ $$-\frac{2}{3} \times m_2 = -1$$ $$m_2 = \frac{-1}{-\frac{2}{3}}$$ $$m_2 = \frac{3}{2}$$

The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Using points A $(2, 1)$ and B $(4, m)$:

$$m_2 = \frac{m - 1}{4 - 2}$$ $$\frac{3}{2} = \frac{m - 1}{2}$$

Multiply both sides by 2:

$$3 = m - 1$$

Add 1 to both sides:

$$m = 3 + 1$$ $$m = 4$$

The value of $m$ is $\boxed{4}$.

The given quadratic equation is $2x^2 - 5x + 3 = 0$.
Divide all terms by 2:

$$x^2 - \frac{5}{2}x + \frac{3}{2} = 0$$

To complete the square for $x^2 + bx$, add $(\frac{b}{2})^2$ to both sides. Here, $b = -\frac{5}{2}$.

$$\left(\frac{b}{2}\right)^2 = \left(\frac{-\frac{5}{2}}{2}\right)^2 = \left(-\frac{5}{4}\right)^2 = \frac{25}{16}$$

Add $\frac{25}{16}$ to both sides of the equation:

$$x^2 - \frac{5}{2}x + \frac{25}{16} = -\frac{3}{2} + \frac{25}{16}$$

The left side is a perfect square: $(x - \frac{5}{4})^2$.
For the right side, find a common denominator:

$$-\frac{3}{2} + \frac{25}{16} = -\frac{3 \times 8}{2 \times 8} + \frac{25}{16} = -\frac{24}{16} + \frac{25}{16} = \frac{1}{16}$$

So the equation becomes:

$$\left(x - \frac{5}{4}\right)^2 = \frac{1}{16}$$

$$x - \frac{5}{4} = \pm\sqrt{\frac{1}{16}}$$ $$x - \frac{5}{4} = \pm\frac{1}{4}$$

$$x = \frac{5}{4} \pm \frac{1}{4}$$

Two possible solutions:

$$x_1 = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2}$$ $$x_2 = \frac{5}{4} - \frac{1}{4} = \frac{4}{4} = 1$$

The solutions are $\boxed{x = \frac{3}{2} \text{ or } x = 1}$.

The given equation is $\text{Log } (x-1) = \text{log } 12 - \text{log } (x-2)$.
Use the logarithm property $\text{log } a - \text{log } b = \text{log } \left(\frac{a}{b}\right)$ on the right side:

$$\text{Log } (x-1) = \text{log } \left(\frac{12}{x-2}\right)$$

Since the logarithms on both sides have the same base (implied base 10), their arguments must be equal:

$$x-1 = \frac{12}{x-2}$$

Multiply both sides by $(x-2)$ to eliminate the denominator:

$$(x-1)(x-2) = 12$$

Expand the left side:

$$x^2 - 2x - x + 2 = 12$$ $$x^2 - 3x + 2 = 12$$

Subtract 12 from both sides to set the quadratic equation to zero:

$$x^2 - 3x - 10 = 0$$

Factor the quadratic equation:
We need two numbers that multiply to -10 and add to -3. These numbers are -5 and 2.

$$(x-5)(x+2) = 0$$

This gives two possible solutions for $x$:

$$x-5 = 0 \implies x = 5$$ $$x+2 = 0 \implies x = -2$$

For a logarithm $\text{log } A$ to be defined, the argument $A$ must be positive ($A > 0$).
Check $x=5$:
$x-1 = 5-1 = 4 > 0$ (Valid)
$x-2 = 5-2 = 3 > 0$ (Valid)
So, $x=5$ is a valid solution.

Check $x=-2$:
$x-1 = -2-1 = -3$ (Not valid, as argument must be positive)
$x-2 = -2-2 = -4$ (Not valid)
So, $x=-2$ is not a valid solution.

The only valid solution is $\boxed{x=5}$.