JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Three Dimensional Geometry

Take a general point on the line as $P=(-3+10\lambda ,2-7\lambda ,\lambda )$, with direction ratios $(10,-7,1)$. The vector from $M(2,-1,4)$ to $P$ is $(-5+10\lambda ,3-7\lambda ,\lambda -4)$, and at the foot of the perpendicular this is orthogonal to the line: $10(-5+10\lambda )-7(3-7\lambda )+(\lambda -4)=0$ gives $150\lambda =75$, so $\lambda =\frac{1}{2}$. Then $P=\left(2,-\frac{1}{2},\frac{1}{2}\right)$ and $|MP|=\sqrt{0+\frac{1}{4}+\frac{49}{4}}=\sqrt{\frac{50}{4}}=\frac{5}{\sqrt{2}}\approx 3.54$, which lies between $3$ and $4$.