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1) Write the vector equation of the line in $\mathbb R^3$ given by:

\begin{align} x+y+z&=6 \\ x+2y+z&=1 \end{align}

2) Write the vector equation of the plane in $\mathbb R^3$ that passing through (1,0,-1) and is perpendicular to the line of exercise 1)

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Subtracting the first equation from the second gives $y=-5$. In this plane, you want all the points satisfying $x+z=11$, which is a line.

Parametrically, we can give it as $(t,-5,11-t)$, for all real $t$.

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For your second question, a plane is defined by a point and a normal direction (a direction that is perpendicular to every vector in your plane). In this exercise, this direction is the direction vector of your line which was given by vadim123: $\vec{d}=(1,0,-1)$. Using that and the given point, the equation of the plane is $$((x,y,z) - (1,0,-1))\bullet (1,0,-1)=0$$ Also, you can write your equation as $$x-z =2$$

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