The distance of the point (1,1,9) from the point of intersection of the line (x-3)/(1)=(y-4)/(2)=(z-5)/(2) and the plane x+y+z=17 is :

3D Geometry: The distance of the point (1,1,9) from the point of intersection of… | JEE Main 2021

JEE Main 2021 (Online) 24th February Morning ShiftQuestion 1

The distance of the point $(1,1,9)$ from the point of intersection of the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ and the plane $x+y+z=17$ is :

3D Geometry

$2\sqrt{19}$$19\sqrt{2}$$\sqrt{38}$✓ Correct answer$38$

Solution

Point $(3+t,4+2t,5+2t)$: $12+5t=17$, $t=1$, point $(4,6,7)$. Distance to $(1,1,9)$: $\sqrt{9+25+4}=\sqrt{38}$.

3D Geometry - JEE Mathematics question (JEE Main 2021 (Online) 24th February Morning Shift)

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