(x-1)/(1)=(y-2)/(-1)=(z-3)/(2) and L2: (x-4)/(2)=y/1=(z+1)/(-1).

Direction vectors $\vec d_1=(1,-1,2)$, $\vec d_2=(2,1,-1)$ give $\vec d_1\times\vec d_2=(-1,5,3)$ with magnitude $\sqrt{35}$. (A) With $\vec a_2-\vec a_1=(3,-2,-4)$, the scalar triple product is $-3-10-12=-25$, so the distance is $25/\sqrt{35}=\frac{5\sqrt{35}}{7}$; correct. (B) The plane through $(1,2,3)$ with normal $(-1,5,3)$ is $x-5y-3z+18=0$ (the constant $+8$ fails at $(1,2,3)$); incorrect. (C) $\cos\theta=\frac{|2-1-2|}{\sqrt6\sqrt6}=\frac16$, not $\frac13$; incorrect. (D) Solving the perpendicularity conditions gives $s=-\tfrac27$, so the foot on $L_1$ is $\left(\tfrac57,\tfrac{16}7,\tfrac{17}7\right)$, and the stated $z$-coordinate is wrong; incorrect. Answer: A only. Always compare the numerators carefully: the plane constant, the angle and the foot of the perpendicular are easy to misstate, which is why only the distance claim survives.