# A line with direction ratios < 2, 2, 1 > intersects the lines $\frac{x-7}{3}= \frac{y-5}{2}= \frac{z-3}{1}\: and\: \frac{x-1}{2}= \frac{y+1}{4}= \frac{z+1}{3}$at the points P and Q respectively. Find the length and the equation of the intercept PQ.

Let a be the common value of $\frac{x-7}{3}= \frac{y-5}{2}= \frac{z-3}{2}= a$
we can rewrite there equation under the following equivalent parametric form
$\begin{pmatrix} x\\ y \\ z \end{pmatrix}= \begin{pmatrix} 7+3a\\ 5+2a \\ 3+2a \end{pmatrix}---(1)$
In the same way, the generic point on the second straight line is
$\begin{pmatrix} x\\ y \\ z \end{pmatrix}= \begin{pmatrix} 1+2b\\ -1+4b \\ -1+3b \end{pmatrix}---(2)$
Then you just have to express the proportionality of $\overrightarrow{PQ}$ with the given, vector, yielding the following system of 2 equations and 2 unknowns.
$\frac{\left ( 1+2b \right )-\left ( 7+3a \right )}{2}= \frac{\left ( -1+4b \right )-\left ( 5+2a \right )}{2}= \frac{\left ( -1+3b \right )\left ( -3+2a \right )}{1}$
giving $a= -\frac{2}{3}\; \; and\: \: b= \frac{1}{3}$
It remains to  placing there values in (1) and (2) to get the coordinates of
$P= \begin{pmatrix} 5\\ \frac{11}{3} \\ \frac{5}{3} \end{pmatrix}\: and\; Q= \begin{pmatrix} \frac{5}{3}\\ \frac{1}{3} \\0 \end{pmatrix}$
The length of PQ
$PQ= \sqrt{\left ( \frac{5}{3}-5 \right )^{2}+\left ( \frac{1}{3}-\frac{11}{3} \right )^{2}+\left ( 0\, -\frac{5}{3} \right )^{2}}$
$= \sqrt{\left ( \frac{-10}{3} \right )^{2}+\left (- \frac{10}{3} \right )^{2}+\left ( -\frac{5}{3} \right )^{2}}$
$= \sqrt{\frac{100}{9}+\frac{100}{9}+\frac{25}{9}}$
$= \sqrt{\frac{225}{9}}= \frac{15}{3}= 5\, units$
The equation of intercept PQ is
$\frac{x-x_{1}}{x_{2}-x_{1}}= \frac{y-y_{1}}{y_{2}-y_{1}}= \frac{z-z_{1}}{z_{2}-z_{1}}$
$\frac{x-5}{\frac{-10}{3}}= \frac{y-\frac{11}{3}}{-\frac{10}{3}}= \frac{z-\frac{5}{3}}{-\frac{5}{3}}$
$\frac{3x-15}{-10}= \frac{3y-11}{-10}= \frac{3z-5}{-5}$   is the required equation

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