Find the distance of the point (2, 3, 4) measured along the line (x -4)/3=(y +5)/6=(z +1)/2 from the plane 3 x +2 y +2 z +5=0. 2 2 3 3 3 3 3 3 [

Equation of line through ($\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$) and parallel to the given line is \[\frac{x-2}{3}=\frac{y-3}{6}=\frac{z-4}{2} \] Any point on this line is \(\displaystyle (3 \lambda+2,6 \lambda+3,2 \lambda+4)\) If it lies on the plane , then \[\begin{gathered} 3(3 \lambda+2)+2(6 \lambda+3)+2(2 \lambda+4)+5=0 \\ \Rightarrow \lambda=-1 \end{gathered} \] Hence, the point is (-$\displaystyle 1$, -$\displaystyle 3$, $\displaystyle 2$) Distance \(\displaystyle =\sqrt{9+36+4}=7\)