Let the point $(-1,\alpha,\beta)$ lie on the line of the shortest distance between the lines $\dfrac{x+2}{-3}=\dfrac{y-2}{4}=\dfrac{z-5}{2}$ and $\dfrac{x+2}{-1}=\dfrac{y+6}{2}=\dfrac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to _____
Step-by-Step Solution
Key Concept: Direction of shortest distance line $=\vec{b_1}\times\vec{b_2}=(-3,4,2)\times(-1,2,0)=(4\cdot0-2\cdot2, 2\cdot(-1)-(-3)\cdot0, -3\cdot2-4\cdot(-1))=(-4,-2,-2)\propto(2,1,1)$.
$\alpha=-3,\beta=2$. $(\alpha-\beta)^2=25$.
Correct Answer: 25