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)