• # question_answer If the line,$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}$meets the plane, $x+2y+3z=15$at a point P, then the distance of P from the origin is [JEE Main 9-4-2019 Morning] A) $\frac{9}{2}$B) $2\sqrt{5}$C) $\frac{\sqrt{5}}{2}$                         D) $\frac{7}{2}$

Any point on the given line can be $(1+2\lambda ,-1+3\lambda ,2+4\lambda );\lambda \in R$ Put in plane $1+2\lambda ,+(-2+6\lambda )+(6+12\lambda )=15$ $20\lambda +5=15$ $20\lambda =10$ $\lambda =\frac{1}{2}$ $\therefore$ Point$\left( 2,\frac{1}{2},4 \right)$ Distance from origin $=\sqrt{4+\frac{1}{4}+16}=\sqrt{\frac{16+1+64}{2}}=\sqrt{\frac{81}{2}}$$=\frac{9}{2}$