what is the projection of point in 3d plane If the coordinate $T(3,4,3) ,B(6,0,0),C(6,0,6)$ and $P(2,0,0)$ in $3$ -dimensions. What is the coordinate of the foot altitude of $P$ with respect to plane $TBC$.
My-Progress:
I am trying to use the equation of plane before resulting
$4x+3y=24$. What I need to do next or this step is unnecessary.

Write the vector parametric equation of the plane.  This is
$r(t,s) = T + t (B - T) + s (C - T) $
Substituting
$r(t,s) = (3,4,3) + t (3, -4, -3) + s(3, -4, 3) $
So the two vectros spanning the plane are $\mathbf{v_1} = [3, -4, -3]^T $ and $\mathbf{v_2} = [3, -4, 3]^T$.
Define the $3 \times 2 $ matrix $A$ as follows
$ A = [\mathbf{v_1, v_2} ] = \begin{bmatrix} 3 && 3 \\ -4 && -4 \\ -3 && 3 \end{bmatrix} $
Then the projector onto the plane is
$P = A (A^T A)^{-1} A^T $
Such the projection of a point $\mathbf{p}$ onto the plane is given by
$ \text{Proj}(\mathbf{p}) = T + P (\mathbf{p} - T) $
Let's compute the projection matrix $P$.  We have
$ (A^T A)^{-1} = \dfrac{1}{900} \begin{bmatrix} 34 && - 16 \\ -16 &&34 \end{bmatrix} $
Therefore,
$P = \begin{bmatrix} 0.36 && -0.48 && 0 \\ -0.48 && 0.64 && 0 \\ 0 && 0 && 1 \end{bmatrix} $
And finally, the projection of $Q (2, 0, 0)$ is
$ \text{Proj}(Q) = (3,4,3) + P (Q - (3,4,3) ) = (3,4,3) + P (-1, -4, -3) \\= (3,4,3) + ( 1.56, -2.08,  -3 ) = (4.56, 1.92, 0) $
