Put the two given equations together in a system, set $x=0$ (say) and solve for $y,z$ to get an actual point on the intersection line. Accompanied by its direction vector and the outlying point $P$, we can determine three points from the desired plane and subsequently determine an equation for it.
Dostre's edit (see comments)
$x=0:\begin{cases} y-2z-3=0 \\ -y+z-2=0 \end{cases}$ ;$\;\;\;\;$ add them up and you get:
$\;\;\;\;\;\;\;\;\;\;\begin{cases} -z-5,\;\;z=-5\\ -y+z-2=0,\;y=-7 \end{cases}\Rightarrow$Point $ (0,-7,-5)$, call it Q, is on the line of intersection.
So now we have two points $P(-1,4,2)$ and $Q(0,-7,-5)$ on our desired plane and vector w thats is parallel to the desired plane. In order to find the equation of the desired plane we need a vector that is normal to it.We can find that normal vector by taking cross product of two vectors that are parallel to the desired plane. We already have w so the other vector will be
*PQ*$<0-(-1),-7-4,-5-2>=<1,-11,-7>$
Now normal vector to desired plane will be the cross product of w and PQ:
PQ x w=$\begin{vmatrix} i & j & k \\ 1 & -11 & -7 \\ 1 & 10 & 6 \end{vmatrix}=i\begin{vmatrix} -11 & -7 \\ 10 & 6 \end{vmatrix}-j\begin{vmatrix} 1 & -7 \\ 1 & 6 \end{vmatrix}+k\begin{vmatrix} 1 & -11\\ 1 & 10 \end{vmatrix}=4i-13j+21k$
$4i-13j+21k=<4,-13,21>$ is the vector normal to the desired plane.
A normal vector of a 2-dimensional line will have the direction vector of an orthogonal line to it.
$$L: 3x-5y=1 \iff y = \dfrac{3}{5}x-\dfrac{1}{5}$$
So any line having slope $-\dfrac{5}{3}$ will be orthogonal to $L$. In other words, all normal vectors to $L$ will be a non-zero multiple of $\langle3,-5\rangle$. In general, a normal to
$$ax+by=c$$
will be a scalar multiple of $\langle a,b\rangle$.
If you're looking for the unique orthogonal line passing through your given $P$ then you use the point in the point-slope equation of a line.
$$y-(-2) = \dfrac{-5}{3}(x-3)$$
Best Answer
Let the line and point have position vectors $\vec r=\vec a+\lambda \vec b$ ($\lambda$ is real) and $\vec p$ respectively. Set $(\vec r-\vec p).\vec b=0$ and solve for $\lambda$ to obtain $\lambda_0$. The normal vector is simply $\vec a+\lambda_0 \vec b-\vec p$.