Given line of intersection of planes $\overrightarrow r.(3\hat i-\hat j+\hat k)=1.......(i)$

and $\overrightarrow r.(\hat i+4\hat j-2\hat k)=2.............(ii)$

Normal to $(i)\;and\:(ii) $ are $\overrightarrow n_1=(3,-1,1)\:\;and\:\:\overrightarrow n_2=(1,4,-2)$

The line of intersection of $(i)\:and\:(ii) $ is $\perp$ to normals to both the planes.

$\therefore$ The line is along $\overrightarrow n_1\times\overrightarrow n_2$

$\overrightarrow n_1\times\overrightarrow n_2=\left |\begin {array}{ccc} \hat i & \hat j & \hat k \\ 3 & -1 & 1 \\1 & 4 & -2\end {array}\right |=(-2,7,13)$

$\therefore$ $d.r.$ of the line of intersection is $(2,-7,-13)$

$\Rightarrow\:$The plane through $(1,2,-1) $ and $\perp$ to the line of intersection will have normal $\overrightarrow n$ =$(2,-7,-13)$

$\therefore$ The eqn. of the required plane is $[\overrightarrow r-(\hat i+2\hat j-\hat k)].\overrightarrow n=0$

$\Rightarrow\:\overrightarrow r.(2\hat i-7\hat j-13\hat k)=1$