Find the shortest distance between the z-axis and the line, $x+y+2z-3=0,2x+3y+4z-4=0.$

Solution: let $p_1$ and $p_2$ be the points on the given planes having coordinates (1,1,2) & (2,3,4) respectively
R = $\left|\begin{array}{ccc}i& j& k\\ 1& 1& 2\\ 2& 3& 4\end{array}\right|$
$=i\left(-2\right)-j\left(0\right)+k\left(1\right)=-2i+k$
$\left(-2,0,1\right)$
if z=0 \displaystyle{x}+{y}-{3}={0}&{2}{x}+{3}{y}-{4}={0}
we get $2\left(3-y\right)+3y-4=0$
then, \displaystyle{y}=-{2}&{x}={5}
so,intersection point $\left(5,-2,0\right)$
eqn of line is
$\frac{x-5}{-2}=\frac{y+2}{0}=\frac{z}{1}$
let direction ratios of this line is $\left(a_1,b_1,c_1\right)$
eqn of z axis is
$\frac{x}{0}=\frac{y}{0}=\frac{z}{0}$
let direction ratios of this line is $\left(a_2,b_2,c_2\right)$
S.D = $\frac{\left|\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|\right|}{\left|\left|\begin{array}{ccc}i& j& k\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|\right|}$
$=\frac{\left|\left|\begin{array}{ccc}-5& 2& 0\\ -2& 0& 1\\ 0& 0& 1\end{array}\right|\right|}{\left|\left|\begin{array}{ccc}i& j& k\\ -2& 0& 1\\ 0& 0& 1\end{array}\right|\right|}$
$\frac{\left|\left(-4\right)\right|}{\left|j\left(2\right)\right|}=2$