If the point C(− 2, 3) is equidistant from the points A(3, − 1) and B(x, 8),

As per the question, we have

$A C=B C$

$\Rightarrow \sqrt{(-2-3)^{2}+(3+1)^{2}}=\sqrt{(-2-x)^{2}+(3-8)^{2}}$

$\Rightarrow \sqrt{(5)^{2}+(4)^{2}}=\sqrt{(x+2)^{2}+(-5)^{2}}$

$\Rightarrow 25+16=(x+2)^{2}+25 \quad$ (Squaring both sides)

$\Rightarrow 25+16=(x+2)^{2}+25$

$\Rightarrow(x+2)^{2}=16$

$\Rightarrow x+2=\pm 4$

$\Rightarrow x=-2 \pm 4=-2-4,-2+4=-6,2$

Now

$B C=\sqrt{(-2-x)^{2}+(3-8)^{2}}$

$=\sqrt{(-2-2)^{2}+(-5)}$

$=\sqrt{16+25}=\sqrt{41}$ units

Hence, $x=2$ or $-6$ and $B C=\sqrt{41}$ units.