The second one, fourth one, and sixth one are true.
Hope this helps!
The normal vector to the plane <em>x</em> + 3<em>y</em> + <em>z</em> = 5 is <em>n</em> = (1, 3, 1). The line we want is parallel to this normal vector.
Scale this normal vector by any real number <em>t</em> to get the equation of the line through the point (1, 3, 1) and the origin, then translate it by the vector (1, 0, 6) to get the equation of the line we want:
(1, 0, 6) + (1, 3, 1)<em>t</em> = (1 + <em>t</em>, 3<em>t</em>, 6 + <em>t</em>)
This is the vector equation; getting the parametric form is just a matter of delineating
<em>x</em>(<em>t</em>) = 1 + <em>t</em>
<em>y</em>(<em>t</em>) = 3<em>t</em>
<em>z</em>(<em>t</em>) = 6 + <em>t</em>
352 would be the answer I think I multiplied it and searched it up to make sure I was right
