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**Soroban** Hello, hiy312!

Two lines are coplanar if they intersect.

We have: .$\displaystyle \begin{Bmatrix}x &=& t+1 \\ y &=& 2t+2 \\ z &=& t-3 \end{Bmatrix}\quad \begin{Bmatrix}x &=& u-1 \\ y &=& u+3 \\ z &=& \frac{u-5}{2}\end{Bmatrix}$

Equate coordinates: .$\displaystyle \begin{Bmatrix} t+1 &=& u-1 && \Rightarrow && t-u &=& -2 \\ 2t + 2 &=& u+3 && \Rightarrow && 2t - u &=& 1 \\ t-3 &=& \frac{u-5}{2} && \Rightarrow && t - \frac{u}{2} &=& \frac{1}{2} \end{Bmatrix}$

Solve the system of equations: .$\displaystyle t = 3,\;u = 5$

The lines intersect at $\displaystyle P(4,8,0)$ . . . They are coplanar.

The normal of the plane is perpendicular to both lines.

The lines have vectors: .$\displaystyle \langle 1,2,1\rangle\,\text{ and }\,\langle 1,1,\tfrac{1}{2}\rangle $

Hence: .$\displaystyle \vec n \;=\;\begin{vmatrix} i&j&k \\ 1&2&1 \\ 1&1&\frac{1}{2}\end{vmatrix} \;=\;i(1-1) - j(\tfrac{1}{2}-1) + k(1-2) \;=\;0i + \tfrac{1}{2}j - k $

. . . . . . $\displaystyle \vec n \;=\;\langle 0,\,\tfrac{1}{2},\,\text{-}1\rangle \;=\;\langle 0,\,1,\,\text{-}2\rangle $

The equation of the plane through $\displaystyle (4,8,0)$ with $\displaystyle \vec n \,=\,\langle 0,1,\text{-}2\rangle$ is:

. . $\displaystyle 0(x-4) + 1(y-8) - 2(z-0) \:=\:0 \quad\Rightarrow\quad y - 2z -8 \;=\;0$