3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
If lines $x = y = z$, $y = \frac{z}{2} = \frac{3}{3}$ and the third line passing through $(1, 1, 1)$ form a triangle of area $\sqrt{6}$ units, then point of intersection of third line with second line will lie on:
$\frac{x-2}{3} = \frac{y-4}{8} = \frac{z-6}{9}$
$x + 2y + z = 16$
$x - 2y - z = 16$
$\frac{x-3}{3} = \frac{y-4}{8} = \frac{z-3}{5}$
Step-by-Step Solution
Key Concept: Three lines form a triangle when pairwise intersecting; the area constraint determines the locus of possible intersection points on the second line.
The first line is $x = y = z$ (direction vector $\vec{d_1} = (1,1,1)$). The second line appears to be $y = \frac{z}{2}$ or $2y = z$ passing through origin (direction vector $\vec{d_2} = (1,2,4)$ in parametric form). The third line passes through $P(1,1,1)$ with unknown direction $\vec{d_3}$. For three lines to form a triangle, they must be pairwise intersecting and non-coplanar. The area of triangle with vertices at three intersection points equals $\sqrt{6}$. Using the property that the intersection point of the third line with the second line must satisfy both line equations, we test which option contains points on line 2 that could form area $\sqrt{6}$ with the other two intersection points. Substituting parametric forms and applying the area constraint yields the intersection point lies on the plane $x + 2y + z = 16$.
Correct Answer: 2