Q.Find the shortest distance between the lines: (x+1)/4 = (y-3)/-6 = (z+1)/1 and (x+3)/3 = (y-5)/2 = (z-7)/6. OR Find the image of the point (5, -3, 1) in the plane 2x - 2y - 3z = 10.

Using the shortest-distance-between-skew-lines formula with the two given lines gives 242/√2561 (≈4.78 units).

Primary question — shortest distance between the two lines:

Line 1: $\dfrac{x+1}{4}=\dfrac{y-3}{-6}=\dfrac{z+1}{1}$, passing through $A_1=(-1,3,-1)$ with direction $\vec b_1=(4,-6,1)$.

Line 2: $\dfrac{x+3}{3}=\dfrac{y-5}{2}=\dfrac{z-7}{6}$, passing through $A_2=(-3,5,7)$ with direction $\vec b_2=(3,2,6)$.

Shortest distance formula for skew lines:

$d = \dfrac{|(\vec A_2-\vec A_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}$

$\vec A_2-\vec A_1 = (-2,2,8)$

$\vec b_1\times\vec b_2 = \begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\4&-6&1\\3&2&6\end{vmatrix}$

$\hat\imath: (-6)(6)-(1)(2)=-36-2=-38$

$\hat\jmath: -[(4)(6)-(1)(3)]=-(24-3)=-21$

$\hat k: (4)(2)-(-6)(3)=8+18=26$

$\vec b_1\times\vec b_2 = (-38,-21,26)$

$(\vec A_2-\vec A_1)\cdot(\vec b_1\times\vec b_2) = (-2)(-38)+(2)(-21)+(8)(26) = 76-42+208=242$

$|\vec b_1\times\vec b_2| = \sqrt{38^2+21^2+26^2}=\sqrt{1444+441+676}=\sqrt{2561}$

$d = \dfrac{242}{\sqrt{2561}} \approx 4.78$ units

…