Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The given line is $2x=3y$ and $z=1$, which can be rewritten as $\frac{x}{3}=\frac{y}{2}=\frac{z-1}{0}$.

The direction ratios of the line are $(3,2,0)$.

The plane passes through the points $A(2,1,2)$ and $B(1,2,1)$.

The direction ratios of the line segment $AB$ are $(1-2,2-1,1-2)=(-1,1,-1)$.

The normal vector $\vec{n}$ to the plane is perpendicular to both the given line and the line segment $AB$.

$\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& 0\\ -1& 1& -1\end{array}\right|=\hat{i}(-2-0)-\hat{j}(-3-0)+\hat{k}(3-(-2))=-2\hat{i}+3\hat{j}+5\hat{k}$

The direction ratios of the normal are $(-2,3,5)$.

The equation of the plane passing through $A(2,1,2)$ is:

$-2(x-2)+3(y-1)+5(z-2)=0$

$-2x+4+3y-3+5z-10=0$

$2x-3y-5z+9=0$

Substituting the point $(-2,0,1)$ into the equation of the plane:

$2(-2)-3(0)-5(1)+9=-4-0-5+9=0$

The point $(-2,0,1)$ satisfies the equation of the plane.

Answer: $(-2,0,1)$