JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Three Dimensional Geometry

Rewrite the line as $\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda }$, so direction ratios are $(1,2,\lambda )$.

The plane $x+2y+3z=4$ has normal vector $(1,2,3)$.

The angle $\theta$ between line and plane satisfies:
$\sin \theta =\frac{|1+4+3\lambda |}{\sqrt{5+{\lambda }^2}\cdot \sqrt{14}}$

Given $\cos \theta =\sqrt{\frac{5}{14}}$, we have $\sin \theta =\sqrt{1-\frac{5}{14}}=\frac{3}{\sqrt{14}}$.

Thus: $\frac{|5+3\lambda |}{\sqrt{5+{\lambda }^2}\cdot \sqrt{14}}=\frac{3}{\sqrt{14}}$

Solving: $|5+3\lambda |=3\sqrt{5+{\lambda }^2}$, which gives $\lambda =\frac{2}{3}$.

Parametric point on line: $(t,2t+1,\frac{2t}{3}+3)$. Substituting into plane equation:
$t+2(2t+1)+3(\frac{2t}{3}+3)=4$
$t+4t+2+2t+9=4$
$7t=-7$, so $t=-1$.

Point: $(-1,-1,\frac{7}{3})$, so $\alpha +2\beta +6\gamma =-1-2+14=11$.