If the lines x – my + 3 = 0 and 2x + 3y – 7 = 0 are perpendicular to each other, then the value of m is:

1. $–\dfrac{2}{3}$

2. $\dfrac{2}{3}$

3. $–\dfrac{3}{2}$

4. $\dfrac{3}{2}$

For two lines to be perpendicular, the product of their slopes must be -1.

Line 1: x - my + 3 = 0

First, convert the equation x - my + 3 = 0 into the slope-intercept form, y = mx + c, to find its slope, m.

$-my = -x - 3 \\[1em] y = \dfrac{-1}{-m}x + \dfrac{-3}{-m} = \dfrac{1}{m}x + \dfrac{3}{m} \\[1em] m_1 = \dfrac{1}{m}.$

Line 2: 2x + 3y - 7 = 0

First, convert the equation 2x + 3y - 7 = 0 into the slope-intercept form, y = mx + c, to find its slope, m.

$3y = -2x + 7 \\[1em] y = -\dfrac{2}{3}x + \dfrac{7}{3} \\[1em] m_2 = -\dfrac{2}{3}$

The product of slopes of perpendicular lines is equal to -1:

$\Rightarrow m$1 \times m2 = -1 \\[1em] \Rightarrow \Big(\dfrac{1}{m}\Big) \times \Big(-\dfrac{2}{3}\Big) = -1 \\[1em] \Rightarrow \dfrac{-2}{3m} = -1 \\[1em] \Rightarrow -2 = -3m \\[1em] \Rightarrow m = \dfrac{2}{3}.

Hence, option 2 is the correct option.