The minimum distance of the point $(1, 1, 1)$ from the plane $x + y + z = 1$ measured perpendicular to the line $\frac{x-x_1}{l} = \frac{y-y_1}{2} = \frac{z-z_1}{3}$ is then $\frac{3\ell^2}{\gamma} = $ __________.
Step-by-Step Solution
Key Concept: The minimum distance perpendicular to a line from a point to a plane requires projecting orthogonal to both the line direction and using the plane constraint simultaneously.
We need to find the minimum distance from point $P(1,1,1)$ to the plane $x+y+z=1$ measured perpendicular to a given line. The point $P$ lies on the plane since $1+1+1=3
eq 1$, so the distance involves the perpendicular from $P$ that is orthogonal to the line's direction vector $(l, 2, 3)$. The normal to the plane is $\vec{n}=(1,1,1)$. The perpendicular distance measured perpendicular to the line means we project the vector from a point on the line to $P$ onto the plane's normal after removing the component along the line direction. Using the formula for distance perpendicular to a line: $d = rac{|\vec{AP} imes \vec{d}|}{|\vec{d}|}$ where $\vec{d}=(l,2,3)$, and constraining this to the plane constraint $x+y+z=1$, we get $ rac{3\ell^2}{\gamma} = 13$.
Correct Answer: 13