AP EAMCET Engineering 2012 Solved Paper

Put r=xi+yj+zk
∴xi+yj+zk=(i−6j+2k) +t(i+2j+k)
∴ Any point on line is P(1+t,−6+2t,2+t) is satisfied the second equation of line.
∴(1+t)i+(−6+2t)j+(2+t)k
=2ui+(4+u)j+(1+2u)k
On equating the coefficients of i,j and k, we get
1+t=2u
⇒ t−2u+1=0 ........(i)
−6+2t=4+u
⇒ 2t−u−10=0 ........(ii)
and 2+t=1+2u
⇒t−2u+1=0 .........(iii)
On solving Eqs. (i) and (ii), we get
t=7,u=4
∴P(1+7,−6+2×7,2+7) =P(8,8,9)