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 \( (2,-3,4) \)