Question

A plane passes through \( (2, 1, 2) \) and \( (1, 2, 1) \) and parallel to the line \( 2x = 3y \) and \( z = 1 \), then the plane also passes through the point

• A. \( (-6, 2, 0) \)
• B. \( (6, -2, 0) \)
• C. \( (-2, 0, 1) \)
• D. \( (2, 0, 1) \)

Hint:

For $ax=by$, direction ratios are $(b, a, 0)$.

Answer 1

The Correct Option is C

Step 1: Concept
A plane equation can be determined using a point and a normal vector perpendicular to two vectors in the plane.

Step 2: Meaning
The vector joining $(2, 1, 2)$ and $(1, 2, 1)$ is $\vec{v_1} = (1, -1, 1)$. The given line $2x=3y, z=1$ has direction ratios $\vec{v_2} = (3, 2, 0)$.

Step 3: Analysis
The normal to the plane $\vec{n} = \vec{v_1} \times \vec{v_2} = (-2, 3, 5)$. The equation is $-2(x-1) + 3(y-2) + 5(z-1) = 0 \implies 2x - 3y - 5z + 9 = 0$.

Step 4: Conclusion
Testing $(-2, 0, 1)$: $2(-2) - 3(0) - 5(1) + 9 = -4 - 5 + 9 = 0$. The point lies on the plane. Final Answer: (C)