<s>[INST] Find the line through $P = (2,4,7) $ that intersects and is perpendicular to the line : $x = -1 + t, y = -2 + 3t, z = 4 $ 
Find the line through $P = (2,4,7) $ that intersects and is
  perpendicular to the line : $x = -1 + t, y = -2 + 3t, z = 4 $

I set up a system of equations and obtained :
$-1 + t = x_0 +2 \\ -2 + 3t = y_0 + 4 \\ 4 = z_0 + 7$
Solving for each and putting them I obtain the vector  $ v_1 = <t -3, 3t-6, -3>$ I then set the direction vector as  $ v_2 =<1,3,4>$ 
Since $v_1 \cdot v_2 = 0 $ means they are orthogonal I obtain :
$<t -3, 3t-6, -3> \cdot <1,3,4> = 0  $
Multiplying out and solving for t I get $t = 10/21$
Would I now just substitute my given value for t back into $t_1$ ?
 [/INST] 1. The plane $P_1$ which includes the line
$$\ell_0: x = -1 + t, y = -2 + 3t, z = 4 $$
and the point $p=(2,4,7)$ is
$$\boxed{3x-y-z=-5}$$
2. The plane $P_2$ perpendicular to the line $\ell_0$ psses through point $p=(2,4,7)$ is
$$\boxed{x+3y=14}$$
3. Intersection of planes $P_1$ and $P_2$ is desired line that is
$$\boxed{\dfrac{x-14}{-3}=y=\dfrac{-z+47}{10}}$$
</s>