Use a plane normal to measure distance and construct a perpendicular plane

$\overrightarrow{AB}=(6,12,8)=2(3,6,4)$, so $l:\mathbf r=(-1,2,3)+\lambda(3,6,4)$. The plane normal is $(2,3,-6)$. Its dot product with the line direction is $6+18-24=0$. But $(-1,2,3)\cdot(2,3,-6)=-14\ne-7$, so the line is parallel and not contained.

Choose $Q=(-7/2,0,0)$ on $p$. Then $\overrightarrow{QA}=(5/2,2,3)$ and the normal magnitude is $7$. The perpendicular distance is $|\overrightarrow{QA}\cdot(2,3,-6)/7|=|5+6-18|/7=1$ unit. The source schematic joins Q to A obliquely and labels the perpendicular height separately; QA itself is not the distance.

The required plane contains $\mathbf d=(3,6,4)$ and $\mathbf n_p=(2,3,-6)$ as directions, hence $\mathbf r=(-1,2,3)+s(3,6,4)+t(2,3,-6)$. Their cross product is $(-48,26,-3)$, giving $\mathbf r\cdot(-48,26,-3)=48+52-9=91$. Thus $-48x+26y-3z=91$. Its normal is orthogonal to the old normal, as required.