1. Write the two given lines in parametric form

A line with direction ratios $(2,1,2)$ meets the two given lines at points $P$ and $Q$.

The first line is $x=y+2=z$ So let x=z=t,
y=t-2 Hence a general point on this line is $P(t)=(t,t-2,t).$

The second line is $x+2=2y=2z$ Let the common value be $s$. Then x=s-2,
y=\frac{s}{2},
z=\frac{s}{2}. So a general point on this line is $Q(s)=\left(s-2,\frac{s}{2},\frac{s}{2}\right).$

2. Use the fact that $\overrightarrow{PQ}$ has direction ratios $(2,1,2)$

Since the required line passes through $P$ and $Q$ and has direction ratios $(2,1,2)$, $\overrightarrow{PQ}=Q-P$ must be parallel to $(2,1,2)$.

Now, $Q-P=\left((s-2)-t,\frac{s}{2}-(t-2),\frac{s}{2}-t\right)$ $=\left(s-t-2,\frac{s}{2}-t+2,\frac{s}{2}-t\right).$

So there exists some scalar $k$ such that $s-t-2=2k \quad ...(1)$ $\frac{s}{2}-t+2=k \quad ...(2)$ $\frac{s}{2}-t=2k \quad ...(3)$

From (2) and (3): $\left(\frac{s}{2}-t+2\right)-\left(\frac{s}{2}-t\right)=k-2k$ $2=-k \Rightarrow k=-2.$

From (3): $\frac{s}{2}-t=2(-2)=-4$ $\frac{s}{2}-t=-4 \Rightarrow s-2t=-8 \quad ...(4)$

From (1): $s-t-2=2(-2)=-4$ $s-t=-2 \quad ...(5)$

Subtract (4) from (5): $(s-t)-(s-2t)=-2-(-8)$ $t=6.$ Then from (5), $s-6=-2 \Rightarrow s=4.$

Therefore, $P=(6,4,6), \qquad Q=\left(2,2,2\right).$

3. Equation of line $PQ$

A direction vector of $PQ$ is $P-Q=(6-2,4-2,6-2)=(4,2,4)=2(2,1,2).$ So line $PQ$ can be written as $\vec r=(2,2,2)+\lambda(2,1,2).$

4. Find the perpendicular distance from $A=(1,2,12)$ to line $PQ$

Take point $A=(1,2,12), \qquad Q=(2,2,2), \qquad \vec d=(2,1,2).$ Then $\overrightarrow{QA}=A-Q=(-1,0,10).$

Distance from a point to a line in 3D is $l=\frac{|\overrightarrow{QA}\times \vec d|}{|\vec d|}.$

Compute the cross product: