MATCH THE FOLLOWING:
(A) The coordinates of a point on the line $x = 4y + 5, z = 3y - 6$ at a distance $3$ from the point $(5, 0, -6)$ is(are)
(B) The plane containing the lines $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 6}{7}$ passes through
(C) A line passes through two points $A(2, -3, -1)$ and $B(8, -1, 2)$. The coordinates of a point on this line nearer to the origin at a distance of $14$ units from $A$ is(are)
(D) The coordinates of the foot of the perpendicular from the point $(3, -1, 11)$ on the line $\frac{x}{2} = \frac{y}{3} = \frac{z - 3}{4}$ is(are)
Step-by-Step Solution
Key Concept: Convert geometric conditions (distances, perpendicularity, collinearity) into algebraic equations using parametric forms and distance formulas.
For (A): Parametrize the line as $(4y+5, y, 3y-6)$ and find $y$ such that distance from $(5,0,-6)$ equals 3. This gives $(4y+5-5)^2 + y^2 + (3y-6+6)^2 = 9$, so $16y^2 + y^2 + 9y^2 = 9$, yielding $26y^2 = 9$, thus $y = ±\frac{3}{\sqrt{26}}$. For (B): Find the plane containing both lines by using two points from each line and the direction vectors to set up the plane equation. For (C): Parametrize line $AB$ as $A + t(B-A) = (2+6t, -3+2t, -1+3t)$ and find point at distance 14 from $A$, where $\sqrt{36t^2 + 4t^2 + 9t^2} = 14$ gives $7t = 14$, so $t = 2$, yielding $(14, 1, 5)$. For (D): Parametrize the line as $(2s, 3s, 3+4s)$ and minimize distance to $(3,-1,11)$: $(2s-3)^2 + (3s+1)^2 + (4s-8)^2$ yields $29s^2 - 80s + 74 = 0$ with $s = 2$, giving foot as $(4, 6, 11)$.
Correct Answer: I need to match each problem (A), (B), (C), (D) with their corresponding answers based on the step-by-step solutions provided.
Let me extract the answers from the solutions:
**(A)** The coordinates of a point on the line at distance 3 from (5,0,-6):
- Solution gives $y = ±\