$\text{ Let }\mathrm{P}\text{ be a plane containing the line }\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+5}{2}\text{ and parallel to the line }$$\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z+5}{7}\text{ . If the point }(1,-1,\alpha )\text{ lies on the plane }P,\text{ then value of }\left|5\alpha \right|\text{ is }$ =

$\begin{array}{l}\text{ Equation of plane is }\left|\begin{array}{ccc}x-1& y+6& z+5\\ 3& 4& 2\\ 4& -3& 7\end{array}\right|=0\\ \text{Expand the determinant: }\\ (x-1)(28+6)-(y+6)(21-8)+(z+5)(-9-16)=0\\ 34x-34-13y-78-25z-125=0\\ 34x-13y-25z=237\\ (1,-1,\alpha )\text{ lies on this plane }\\ 34\left(1\right)-13(-1)-25\alpha =237\Rightarrow -25\alpha =190\Rightarrow \left|5\alpha \right|=38\end{array}$