# Show that the lines   and intersect.Also, find the coordinates of the point of intersection. Find the equation of the plane containing the two lines

Given lines:

$\frac{x-2}{1}= \frac{y-2}{3}= \frac{ z-3}{1}=k$

Point on line: $(2+k,2+3k,k+3)$

$\frac{x-2}{1}= \frac{y-3}{4}= \frac{ z-4}{2}=l$

Point on line: $(2+l,3+4l,4+2l)$

At intersect point

$(2+l,3+4l,4+2l)\equiv (2+k,2+3k,k+3)$

$2+l=2+k \Rightarrow l=k$

$\\3+4l= 2+3k \\ \text{put k=l} \Rightarrow 3+4l= 2+3l \\$

$k=l=-1$

Point of intersection $(1,-1,2)$.

Equation of lien passing through two lines:

$\left|\begin{array}{ccc} x-2 & y-2 & z-3 \\ 1 & 3 & 1 \\ 1 & 4 & 2 \end{array}\right|=0$

$\\ (x-2)(6-4) - (y-2)(2-1) +(z-3)(4-3)=0\\ 2x -4 -y+2+z-3 = 0 \\ 2x-y+z = 5$

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