Step 1:

The equation of the plane containing the given point is

$A(x-1)+B(y-1)+C(z+1)=0$--------(1)

Applying the condition of perpendicularity to the plane (1) with the given planes $x+2y+3z-7=0$ and $2x-3y+4z=0$

$A+2B+3C=7$ and $2A-3B+4C=0$

Step 2:

Solving the equation we get,

$\large\frac{A}{\begin{vmatrix}2 & 3\\-3 & 4\end{vmatrix}}=\large\frac{B}{\begin{vmatrix}3 & 1\\4 & 2\end{vmatrix}}=\large\frac{C}{\begin{vmatrix}1 & 2\\2 & -3\end{vmatrix}}$

$\Rightarrow \large\frac{A}{8+9}=\frac{B}{6-4}=\frac{C}{-3-4}$

$\Rightarrow \large\frac{A}{17}=\frac{B}{2}=\frac{C}{-7}$$=\lambda$(say)

$\therefore A=17\lambda,b=2\lambda$ and $C=-7\lambda$

Step 3:

Substituting these values in equ(1) we get,

$17\lambda(x-1)+2\lambda(y-1)-7\lambda(z+1)=0$

$\Rightarrow 17x+2y-7z=26$

This is the required solution of the plane.