Foot of perpendicular drawn from the origin to the plane \( 2 x-3 y+4 z=29 \) is

Given equation of plane
\[ 2 x-3 y+4 z=29 \rightarrow(1) \]
Normal vector of this plane is given by
\[ \vec{N}=2 \hat{i}-3 \hat{j}+4 \hat{k} \]
Now, equation of line which passes through the origin and has the same direction as normal vectors \( N \rightarrow \) is
\[ \begin{array}{l} \frac{x}{2}=\frac{y}{-3}=\frac{z}{4}=\lambda \rightarrow(2) \\ \Rightarrow x=2-x, y=-3 \lambda, z=4 \lambda \rightarrow(3) \end{array} \]
The coordinates of the foot of the perpendicular is
\[ \begin{array}{l} 4 \lambda+9 \lambda+16 \lambda=29 \\ \Rightarrow 29 \lambda=29 \Rightarrow \lambda=1 \end{array} \]
Putting the value of \( \lambda \) in Eq. (3), we have
\( x=2, y=-3, z=4 \) or \( (2,-3,4) \)