JEE Main 2021 · 24 Feb · Shift 1 · Mathematics Q77

MathematicsSingle correct+4 / -1Question 77 of 90

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

2\sqrt{19}

19\sqrt2

\sqrt{38}

Answer and worked solution

Official answer: C

Intersection. $(3+t)+(4+2t)+(5+2t)=17$ gives $t=1$ : (4, 6, 7). Distance from (1, 1, 9): $\sqrt{9+25+4}=\sqrt{38}$ .

(C)

Topics:Equation of a line in space Angles and distances

JEE Main 2021 · 24 Feb · Shift 1, Mathematics question 77