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What is the symmetric equation of the line passing through the point $P=(2,-3,0)$ that is parallel to the vector ${\bf v}=\langle 3,-1,-4 \rangle$?

This line has no symmetric equation because one of the coordinates of $P$ is $0$.

$\cfrac{x-2}{3} =\cfrac{y+3}{-1} =\cfrac{z}{-4}$

$x=2+3t,\quad y=-3-t,\quad z=-4t$

$3(x-2)-(y+3)-4z=0$

$\cfrac{x-3}{2} =\cfrac{y+1}{-3} =\cfrac{z}{-4}$