Third dimensional theory(School maths)

Question

I took these three vectors (-1,3,3) ;(1,2,3);(x,y,0) and because three vectors will be collinear as line intersect XY plane, therefore, the value of determinant of these three vectors are going to be zero from which I got the condition that x=-2y and according to that option (a) is true but answer is option d . Where am I wrong

Answer 1

A point lies in the $xy$-plane iff its third coordinate is zero. Here subtract the direction vector once from the given point to get $(-2,1,0)$.

Answer 2

you have $$\vec{x}=[-1,3,3]+s[1,2,3]$$ for $$z=0$$ we get $s=-1$

Answer 3

You got that $x=-2y$. This is a line, with infinitely many solutions. But only one of them can be true for this point of intersection, since the line crosses the plane in just one place. To find which one is true, as others have pointed out, you solve $$\begin{pmatrix}\tilde x\\\tilde y\\0\end{pmatrix}=\begin{pmatrix}-1\\3\\3\end{pmatrix}+t\begin{pmatrix}1\\2\\3\end{pmatrix}$$Sure enough you get $\tilde x=-2,\tilde y=1$. This satisfies your equation $x=-2y$, but is not the point of intersection. The actual point of intersection is not in the list, so the answer is (d).