The question asks for the vector equation for the line of intersection by the planes x+2y-2z=5 and 6x-3y+2z=8

Answer is given as

$\displaystyle

\vec r = \left\langle {\frac{{31}}{{15}},\frac{{22}}{{15}},0} \right\rangle + t\left\langle {2,14,15} \right\rangle

$

But when I do the working

Labeling x+2y-2z=5 as equation 1

and 6x-3y+2z=8 as equation 2

Multiply equation 1 by 6 to get 6x+12y-12z=30 call this equation 3

equation 2 minus equation 3 to get

$\displaystyle

\begin{array}{l}

15y = 22 - 14z \\

y = \frac{{22}}{{15}} - \frac{{14}}{{15}}z \\

\end{array}

$

Let z=t

$\displaystyle

y = \frac{{22}}{{15}} - \frac{{14}}{{15}}t

$

Than from equation 1

$\displaystyle

\begin{array}{l}

x + 2y - 2z = 5 \\

x = 5 - 2\left( {\frac{{22}}{{15}} - \frac{{14}}{{15}}t} \right) - 2t \\

x = 5 - \frac{{44}}{{15}} + \frac{{28}}{{15}}t - 2t \\

x = \frac{{31}}{{15}} - \frac{2}{{15}}t \\

\end{array}

$

and so

$\displaystyle

\begin{array}{l}

\vec r = \left\langle {\frac{{31}}{{15}},\frac{{22}}{{15}},0} \right\rangle + t\left\langle {\frac{{ - 2}}{{15}},\frac{{ - 14}}{{15}},1} \right\rangle \\

\vec r = \left\langle {\frac{{31}}{{15}},\frac{{22}}{{15}},0} \right\rangle + t\left\langle { - 2. - 14,15} \right\rangle \\

\end{array}

$

Am I right with my negative signs in <-2,-14,15>

and the answer in the text of <2,14,15> is wrong?