Finding point of intersection using cartesian equation?

Question

L1 is defined by $ \dfrac{x+1}{2} = \dfrac{y-3}{3} = 1-z$

So we can write this as $((2t-1), (3t+3), (1-t)) = r $

L2 passes through $(5,4,2)$ and intersects with L1 at right angles.

I am asked to determine the point of intersections between those two lines.

I have tried to find the dot product of the directional vector of L1 and another directional vector (x,y,z) So I got this $2x+3y-z = 0$ and said that $ x = 3, y=2$ and $ z=12$

So now L2 has the following equation $ r = ((5+3t), (4+2t), (2+12t))$ However when i equate L1 and L2 to find $t$. The equations aren't satisfied, meaning that $t$ isn't common.

Answer 1

The point of intersection lies on $L_1$, so we can express it as $P(2t-1,3t+3,1-t)$ for some $t\in\Bbb R$. We require that the vector from $P\to(5,4,2)$ be orthogonal to the vector $(2,3,-1)$, which is colinear with the line. That is,$$(2t-1,3t+3,1-t)\to(5,4,2)\perp(2,3,-1)$$This implies,$$(2t-6,3t-1,-1-t)\cdot(2,3,-1)=0$$giving the value of $t=1$ and the required point as $(1,6,0)$.

Answer 2

Let $L_2$ $$\vec{x}=[5,4,2]+s[a_x,a_y,a_z]$$ Then we have the condition $$2a_x+3a_y+4a_z=0$$ (by the dot-product) and the system $$2t-sa_x=6$$ $$3t-sa_y=1$$ $$-t-sa_z=1$$ Can you solve this?