The line
[tex](x,y,z)=(11,-8,4)+t(3,-1,1)[/tex]
runs in the direction of its tangent vector; we can get it by taking its derivative:
[tex]\vec T=\dfrac{\mathrm d}{\mathrm dt}\left[(11,-8,4)+t(3,-1,1)\right]=(3,-1,1)[/tex]
Any line that runs perpendicular to this line will have a tangent vector that is orthogonal to [tex]\vec T[/tex] above. So construct some vector [tex]\vec v[/tex] that satisfies this.
[tex]\vec v=(x,y,z)[/tex]
[tex]\vec v\cdot\vec T=(x,y,z)\cdot(3,-1,1)=3x-y+z=0[/tex]
Suppose [tex]z=0[/tex]; then [tex]3x=y[/tex], and we can pick any two values that satisfy this condition. For instance,
[tex]\vec v=(1,3,0)[/tex]
And of course,
(1, 3, 0) • (3, -1, 1) = 3 - 3 + 0 = 0
so [tex]\vec v[/tex] and [tex]\vec T[/tex] are indeed orthogonal.
Now, the line running in the direction of [tex]\vec v[/tex] and passing through the origin can be obtained by scaling
[tex](x,y,z)=(4,5,5)+t(1,3,0)[/tex]
More generally, if you have a direction/tangent vector [tex]\vec v[/tex] and some point [tex]\vec p[/tex], the line through [tex]\vec p[/tex] is given by
[tex](x,y,z)=\vec p+t\vec v[/tex]
