3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
The coordinate of the points on the line $\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ which are at a distance $3\sqrt{2}$ from the point $(1, 2, 3)$
$(-2, -1, 3)$
$(2, 2, 4)$
$\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$
$\left(\frac{47}{11}, \frac{42}{11}, \frac{56}{11}\right)$
Step-by-Step Solution
Key Concept: Parametrize the line and substitute into the sphere equation to find intersection points.
Any point on the given line is $(3\lambda - 2, 2\lambda - 1, 2\lambda + 3)$. Substituting into the sphere equation $(3\lambda - 3)^2 + (2\lambda - 3)^2 + (2\lambda)^2 = 18$ gives $(3\lambda)^2 + (2\lambda - 3)^2 + (2\lambda)^2 = 18$. Solving yields $\lambda = 0$ and $\lambda = \frac{30}{17}$, giving points $(-2, -1, 3)$ and $\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$.
Correct Answer: 1,3