JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Three Dimensional Geometry

Take a point $(\lambda ,6-2\lambda ,5\lambda -8)$ on the given line. Matching it with $(4\mu +5,3\mu +7,\mu -2)$ on the second line gives $\lambda =1$, so $A=(1,4,-3)$. Matching it with $(6\gamma -3,3-3\gamma ,\gamma +6)$ on the third line gives $\lambda =3$, so $B=(3,0,7)$. The mid-point of $AB$ is $(2,2,2)$, and its distance from $2x-2y+z=14$ is $\frac{|2(2)-2(2)+2-14|}{\sqrt{4+4+1}}=\frac{12}{3}=4$. The required distance is $4$.