If the line passing through the points  $(a,2,-4)$  and  $(5,3,b)$ crosses the ZX – plane at the point $(-a+2b,0,a+b)$ , then  $14a+7b$

Equation of line passing through the points $(a,2,-4)$  and  $(5,3,b)$ is given by
 $\frac{x-a}{5-a}=\frac{y-2}{3-2}=\frac{z+4}{b+4}$
 $\Rightarrow \frac{x-a}{5-a}=\frac{y-2}{1}=\frac{z+4}{b+4}\mathrm{........}\left(i\right)$
The equation of ZX-plane,  $y=0$
Since, line (i) meets the ZX-plane.
 $\Rightarrow \frac{x-a}{5-a}=\frac{0-2}{1}=\frac{z+4}{b+4}$
$\Rightarrow \frac{x-a}{5-a}=-2$   and  $\frac{z+4}{b+4}=-2$
  $\Rightarrow x-a=-2(5-a)$  and  $z+4=-2(b+4)$
 $$\begin{gathered} \Rightarrow x=a-10+2aandz+4=-2b-8 \\ \Rightarrow x=3a-10andz=-2b-12andy=0 \\ \therefore 3a-10=-a+2band-2b-12=a+b \\ \Rightarrow 4a-2b=10anda+3b=-12\mathrm{........}\left(ii\right) \\ \Rightarrow 2a-b=5\mathrm{........}\left(iii\right) \end{gathered}$$
On solving Equation (i) and (ii), we get
$$\begin{gathered} a=\frac{3}{7}andb=\frac{-29}{7} \\ \therefore 14a+7b=14\left(\frac{3}{7}\right)+7\left(\frac{-29}{7}\right) \\ =6-29=-23 \end{gathered}$$