The area $B$ of the base rectangular triangle $\Delta abc$ is $B=\frac 12\cdot ab\cdot ac = \frac 12 \sqrt 2\cdot 1$.
The volume of the pyramid is $V=\frac 13 Bh$, where $B$ is the area of the base, and $h$ is the height perpendicular to the base.
The normal vector $\vec n$ of the plane can be deduced from its equation $-x+z=0$, meaning it is $\vec n=(-1,0,1)$.
The height $h$ of the pyramid is the projection of the vector $\overrightarrow{ad}$ onto the normal vector $\vec n$.
The formula for that projection is $h=\frac{\overrightarrow{\mathstrut ad} \cdot \overrightarrow{\mathstrut n}}{\|\vec n\|}$.
So we have:
$$\begin{cases}B=\frac 12\sqrt 2 \\
V=\frac 13 Bh = 4 \\
\vec n = (-1,0,1) \\
h=\frac{\overrightarrow{\mathstrut ad} \cdot \overrightarrow{\mathstrut n}}{\|\vec n\|} = \frac{(-1,1,1)t \cdot (-1,0,1)}{\|(-1,0,1)\|} = \frac{2}{\sqrt 2}t=t\sqrt 2
\end{cases}
\implies V = \frac 13 \cdot \frac 12\sqrt 2 \cdot t\sqrt 2 = 4
\implies t = 12
$$
So point $d$ is $(0,0,0)+(-1,1,1)12=(-12,12,12)$.