$3D$ state of stress at point $(1, 1, -2)$ within a body relative to coordinate axis $(x_1, x_2, x_3)$ is given by
$$ \begin{matrix} 2 & 3.5 & 2.5 \\ 3 & 0 & -1.5 \\ 2.5 & -1.5 & 1 \\ \end{matrix} MPa $$
Asked to determine the normal and shear stresses at the point and on the surface of an internal sphere whose equation is $$ x_1^2 + (x_2 - 2)^2 + x_3^2 = 6$$
I have the normal $$ \frac {1} {\sqrt 6} (e_1 - e_2 -2e_3)$$
solving I get $$ t_n = \frac {1} {\sqrt 6} (-6.5e_1 + 6.5e_2 + 2e_3) $$
and I get $$ \sigma_n = -\frac {17} { 6} $$
How do I find $$ \sigma_s $$ the answer for it is supposed to be $8.67$ MPa but I don't seem to get that.
Edit:
So for the stress, I used
$$ \sigma_s = \sqrt{ |t(n)|^2 - \sigma_n^2} $$
$$ |t(n)^2| = \frac {1} {\sqrt 6} \sqrt{((-6.5)^2 + 6.5^2 + 2^2)} = 3.8405 $$
SO,
$$ \sigma_s = \sqrt{((3.8405)^2 - (- \frac {17}{6} )^2 } = 2.5927 MPa $$
But the answer is supposed to be $8.67$ MPa