The equation of the line passing through the point $M(1,1,1)$ intersect at right angle to the line of intersection of the planes $x+2y-4z=0$  and $\frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c}$ is  then  $⟨a,b,c⟩$are proportional to

Find the symmetric form of the line of intersection of two planes. 

Observing the plane equations, origin is common point to both planes, so that the line of intersection is passing though the origin

The line of intersection of two planes is along the vector  $\overline{n}=\left|\begin{array}{ccc}i& j& k\\ 1& 2& -4\\ 2& -1& 2\end{array}\right|=i\left(0\right)-j\left(10\right)+k(-5)$

The direction ratios of line are proportional to  $⟨0,-10,-5⟩$

Hence the equation of the line is  $\frac{x}{0}=\frac{y}{2}=\frac{z}{1}$and general point on this line is $P(0,2k,k)$

Given that the line passing through the points  $M(1,1,1)$ and  $P(0,2k,k)$ makes right angle with the line of intersection of the given two planes 

The direction ratios of the line joining  $M(1,1,1)$ and $P(0,2k,k)$ are$⟨-1,2k-1,k-1⟩$

Since both lines are intersect at right angle 

 $\begin{array}{c}0(-1)+2(2k-1)+1(k-1)=0\\ 4k-2+k-1=0\\ 5k=3\\ k=\frac{3}{5}\end{array}$

Therefore, the direction ratios of the new line are  $⟨-1,\frac{1}{5},-\frac{2}{5}⟩=⟨-5,1,-2⟩$