The line $l_1$ passes through the point $(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. Then the shortest distance between the line $l_1$ and the line $\frac{x+1}{2} = \frac{y+4}{-3} = \frac{z}{2}$ is:
Step-by-Step Solution
Key Concept: $l_1$ has direction $(2,1,-2)$ (normal to plane). Apply skew lines SD formula.
$\vec{b_1}\times\vec{b_2} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-2\\2&-3&2\end{vmatrix} = (-4,-8,-8) = -4(1,2,2)$. $|\vec{b_1}\times\vec{b_2}|=12$. $SD = \frac{|(-3,-10,-2)\cdot(1,2,2)|}{3} = \frac{|-3-20-4|}{3}=9$. Answer: (4)
Correct Answer: 9