Equation of line passing through the points $(a, 2,-4)$ and $(5,3, b)$ is given by
$\frac{x-a}{5-a}=\frac{y-2}{3-2}=\frac{z+4}{b+4}$
$\Rightarrow \quad \frac{x-a}{5-a}=\frac{y-2}{1}=\frac{z+4}{b+4}$$\ldots(i)$
The equation of $Z X$-plane, $y=0$ Since, line (i) meets the $Z X$-plane.
$\Rightarrow \quad \frac{x-a}{5-a}=\frac{0-2}{1}=\frac{z+4}{b+4}$
$\Rightarrow \quad \frac{x-a}{5-a}=-2$ and $\frac{z+4}{b+4}=-2$
$\Rightarrow \quad x-a=-2(5-a)$ and $z+4=-2(b+4)$
$\Rightarrow \quad x=a-10+2 a$ and $z+4=-2 b-8$
$\Rightarrow \quad x=3 a-10$ and $z=-2 b-12$ and $y=0$
$\therefore \quad 3 a-10=-a+2 b$ and $-2 b-12=a+b$
$\Rightarrow \quad 4 a-2 b=10$ and $a+3 b=-12$ $\ldots(ii)$
$\Rightarrow \quad 2 a-b=5$ $\ldots(iii)$
On solving Eqs (i) and (ii), we get
$a=\frac{3}{7} \text { and } b=\frac{-29}{7}$
$\therefore \quad 14 a+7 b=14\left(\frac{3}{7}\right)+7\left(\frac{-29}{7}\right)=6-29=-23$