Equation of plane containing the line $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ at minimum possible distance from $(-1,-3,1)$ is
Step-by-Step Solution
Key Concept: Plane through line: $a(x-1)+b(y-2)+c(z-3)=0$ with $2a+3b+4c=0$. Point $(-1,-3,1)$ on line: confirm it's not on the line ($x=1+2t=-1\Rightarrow t=-1$; $y=2-3=−1\neq-3$). Distance from $(-1,-3,1)$ to plane min when the vector from the point on line to $(-1,-3,1)$ is perpendicular to plane's direction in the constraint surface.
$7x-2y-2z+3=0$.
Correct Answer: (A) $7x-2y-2z+3=0$