Question

If the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+1}{4}$ meets the plane $x+2y+3z=14$ at $P$, then the distance between $P$ and the origin is

• A. $\sqrt{14}$
• B. $\sqrt{15}$
• C. $\sqrt{13}$
• D. $\sqrt{12}$
• E. $\sqrt{17}$

Answer 1

The correct option is A $\sqrt{14}$

Given equation of line is
$\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+1}{4}$
Let the coordinates of point $P$ is
$(2\lambda -1,3\lambda -1,4\lambda -1)$
line meets the plane $x+2y+3z=14$
$\therefore 2\lambda -1+2(3\lambda -1)+3(4\lambda -1)=14$
$2\lambda -1+6\lambda -2+12\lambda -3=14$
$\Rightarrow \lambda =1$
$\therefore$ Point $P$ is $(2-1,3-1,4-1)=(1,2,3)$
Distance between point $(1,2,3)$ and $(0,0,0)$ is
$d=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$