Animated Solution for Mathematics - Three Dimensional Geometry: If the shortest distance between the lines $\frac{x-\lambda }{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda |$ is equal to _________

The geometry of skew lines in 3D space is defined by their unique direction vectors and points of passage. We are given two lines, $L_1$ and $L_2$, and we must determine the parameter $\lambda$ such that the shortest distance between them is exactly $\frac{44}{\sqrt{30}}$.

For $L_1$, given by $\frac{x-\lambda }{3}=\frac{y-2}{-1}=\frac{z-1}{1}$, the point is ${\vec{a}}_1=(\lambda ,2,1)$ and the direction vector is ${\vec{b}}_1=(3,-1,1)$.

For $L_2$, given by $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$, the point is ${\vec{a}}_2=(-2,-5,4)$ and the direction vector is ${\vec{b}}_2=(-3,2,4)$.

The shortest distance between two skew lines is the length of the segment perpendicular to both. We find this direction using the cross product $\vec{n}={\vec{b}}_1\times {\vec{b}}_2$:

The magnitude of this normal vector is:

We define the vector connecting the two lines as ${\vec{a}}_1-{\vec{a}}_2=(\lambda +2,7,-3)$. The shortest distance $d$ is given by the projection formula:

Substituting our values, we calculate the dot product:

Setting the distance equal to the given value:

Canceling $\sqrt{30}$ from both sides and simplifying the fraction:

This yields two possible linear equations:

The problem asks for the largest possible value of $|\lambda |$. Comparing $|\lambda =-43|=43$ and $|\lambda =1|=1$, the final answer is 43.