I need to use the triple cross product to find a vector that lies in the plane determined by
the point (1, 0, 2) and the line $\frac {x}{2}=\frac {y+1}{3}=\frac {z-2}{-1}$ , and is perpendicular to the line
2026-03-25 23:11:07.1774480267
Find a vector that lies in the plane determined by a line and a point and is perpendicular to that line
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Hint:
Let $A(1,0,2)$. You can simply determine the coordinates of the intersection $H$ of the line and the plane through $A$ perpendicular to the line: you'll then can take the vector $\overrightarrow{AH}$.
As the line has $\;\vec u(2,3,-1)$ as a directing vector, an equation of the plane through $A$ perpendicular to the line is $$2x+3y-z=2\cdot 1+3\cdot 0-1\cdot 2=0.$$ Can you continue?