The tangent vector for the given line is
<em>T</em><em>(t)</em> = d/d<em>t</em> ⟨-2 - 3<em>t</em>, 1 - 4<em>t</em>, -5<em>t</em>⟩ = ⟨-3, -4, -5⟩
On its own, this vector points to a single point in space, (-3, -4, -5).
Multiply this vector by some scalar <em>t</em> to get a whole set of vectors, essentially stretching or contracting the vector ⟨-3, -4, -5⟩. This set is a line through the origin.
Now translate this set of vectors by adding to it the vector ⟨-2, -4, 0⟩, which correspond to the given point.
Then the equation for this new line is simply
<em>L</em><em>(t)</em> = ⟨-3, -4, -5⟩<em>t</em> + ⟨-2, -4, 0⟩ = ⟨-2 - 3<em>t</em>, -4 - 4<em>t</em>, -5<em>t</em>⟩
