JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Three Dimensional Geometry

Parametric form of the line: $x=3+t$, $y=4+2t$, $z=5+2t$.

Substitute into the plane equation:



Point of intersection: $(3+1,4+2,5+2)=(4,6,7)$.

Distance from $(1,1,9)$ to $(4,6,7)$:


Wait, $\sqrt{38}\ne 2\sqrt{19}$. Let me recalculate: $\sqrt{38}=\sqrt{38}$, but $2\sqrt{19}=\sqrt{4\cdot 19}=\sqrt{76}$.

Actually, $\sqrt{38}$ is the correct distance. However, checking the options, option (B) is $2\sqrt{19}=\sqrt{76}$. Let me verify the calculation again.

Distance: $\sqrt{3^2+5^2+(-2{)}^2}=\sqrt{9+25+4}=\sqrt{38}$.

The answer should be $\sqrt{38}$, which is option (A). But if the source answer is (B), there may be an error in the source.