Answer:
The stress tensor in matrix form is:
$$
\begin{bmatrix}
\sigma_{xx} & \tau_{xy} & \tau_{zx} \\
\tau_{xy} & \sigma_{yy} & \tau_{yz} \\
\tau_{zx} & \tau_{yz} & \sigma_{zz}
\end{bmatrix}
=
\begin{bmatrix}
60 & -40 & 30 \\
-40 & -40 & 20 \\
30 & 20 & 20
\end{bmatrix} \text{MPa}
$$
To find the normal and shear stresses on a plane with direction cosines $(0.429, 0.514, 0.743)$, we need to calculate the stress components in that direction. We can use the formula:
$$
\begin{bmatrix}
\sigma_n \\
\tau_{shear}
\end{bmatrix}
=
\begin{bmatrix}
cos(n,x)^2 & cos(n,y)^2 & cos(n,z)^2 & 2cos(n,x)cos(n,y) & 2cos(n,y)cos(n,z) & 2cos(n,z)cos(n,x) \\
\frac{1}{2}sin(2\theta_{xy}) & \frac{1}{2}sin(2\theta_{yz}) & \frac{1}{2}sin(2\theta_{zx}) & cos(\theta_{xy})sin(\theta_{xy}) & cos(\theta_{yz})sin(\theta_{yz}) & cos(\theta_{zx})sin(\theta_{zx})
\end{bmatrix}
\begin{bmatrix}
\sigma_{xx} \\
\sigma_{yy} \\
\sigma_{zz} \\
\tau_{xy} \\
\tau_{yz} \\
\tau_{zx}
\end{bmatrix}
$$
where $\theta_{xy}$, $\theta_{yz}$, and $\theta_{zx}$ are the angles between the plane and the $xy$, $yz$, and $zx$ planes, respectively.
Plugging in the given values, we get:
$$
\begin{bmatrix}
\sigma_n \\
\tau_{shear}
\end{bmatrix}
=
\begin{bmatrix}
0.184 & 0.264 & 0.953 & 0.436 & 0.435 & -0.786 \\
-51.7 & 0 & 29.7 & 20 & 123.7 & -32.2
\end{bmatrix}
\begin{bmatrix}
60 \\
-40 \\
20 \\
-40 \\
20 \\
30
\end{bmatrix}
=
\begin{bmatrix}
23.5 \\
-49.3
\end{bmatrix} \text{MPa}
$$
Therefore, the normal stress on the plane is 23.5 MPa, and the shear stress on the plane is -49.3 MPa.