A plane passing through the intersection of planes $H_1=0$ and $H_2=0$ belongs to the pencil generated by these planes, i.e. it has equation $\lambda H_1+\mu H_2=0$ for some $\lambda, \mu$.
In the present case, the equation will be
$$\lambda(x-z)+\mu(y+4z)=\lambda x +\mu y+ (4\mu-\lambda)z=3\lambda+2\mu, $$
and it will be perpendicular to the plane $x-y+3z=5$ if they have orthogonal normal vectors.
As a vector normal to the plane is $(\lambda, \mu, 4\mu-\lambda)$, the orthogonality condition is
A solution (defined within a non-zero factor) is $\lambda=11$, $\mu=2$, whence the equation
$$ 11x+2y-3z=37. $$