How to determine the equation of plane through origin?

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How to determine the equation of plane which passes through the origin and contains the line x = 2+3t , y = 1-4t, z= 6-t.

I set t = 1 and t = 0 to find the another point, so I can get (0,0,0), (5,-3,5) and (2,1,6)

then do the next step

Is it correct ?

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2 Answers

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The plane is parallel to $(3,-4,-1)$ and $(2,1,6)-(0,0,0)=(2,1,6)$. So

$$(3,-4,-1)\times(2,1,6)=(23,20,-11)$$

is normal to the plane.

The plane is $23x+20y-11z=0$.

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The directing vector of the line in the plane is $\vec v(3,-4,1)$

A point on the line is $A(2,1,6)$

Now get the vector $\vec{OA}(2,1,6)$

Do the vector product $\vec v ×\vec{OA}$ and you will get a normal vector to the plane which is $\vec n_P=\vec v ×\vec{OA}= (-25,-16,11)$

The equation of the plane will then be$$-25x-16y+11z+r=0$$Substitute the coordinates of $A$ in this equation and you'll get that $r=0$

Finally, the equation of the plane (P) is:$$-25x-16y+11z=0$$Or equally,$$25x+16y-11z=0$$

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