Triangle And Line

Relevant wiki: 3D Coordinate Geometry - Problem Solving

An equation of the plane is $ax+by+cz+d=\frac{x}{14}+\frac{y}{21}+\frac{z}{42}-1=0$ . The point-plane distance formula gives $D=\frac{|d|}{\sqrt{a^2+b^2+c^2}}=\left(\frac{1}{14^2}+\frac{1}{21^2}+\frac{1}{42^2}\right)^{-1/2}\approx \boxed{11.225}$

This can also be solved using formula for trirectangular tetrahedron

The altitude $h$ satisfies:

$\frac{1}{h^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$ $\frac{1}{h^2}=\frac{1}{14^2}+\frac{1}{21^2}+\frac{1}{42^2} \Rightarrow h \approx \boxed{11.225}$

A more elementary solution: the plane of $T$ has the equation $3x + 2y + z = 42.$ The normal/perpendicular direction to $T$ is therefore described by the vector $\vec n = (3,2,1)$ . Line $\ell$ consists of all points $(3\lambda,2\lambda,\lambda)$ .

To find the intersection, substitute the equation for $\ell$ into that for $T$ : $3(3\lambda) + 2(2\lambda) + \lambda = 42 \\ 14\lambda = 42 \\ \lambda = 3$ so that the intersection is $(x_P,y_P,z_P) = (9, 6, 3)$ . The distance $OP$ is $\sqrt{9^2+6^2+3^3} = 3\sqrt{14} \approx \boxed{11.225}$