The angle between the two planes x+y-z=3 and 3 x+2 y+z=5 is :

for plane $ax + by + cz + d = 0$

normal to it is $a \hat{i}+b \hat{j} + c \hat{k}$

for $p_1: x+y-z=3$     $\vec{n}_1=\hat{i}+\hat{j}-\hat{k}$

$p_2: 3 x+2 y+z=5$    $\vec{n}_2=3 \hat{i}+2 \hat{j}+\hat{k}$

so angle between p₁ and p₂

= angle between $\vec{n}_1$ and $\vec{n}_2$

so $\vec{n}_1 . \vec{n}_2 = |\vec{n}_1| |\vec{n}_2| \cos \theta$

$\Rightarrow (\hat{i}+\hat{j}-\hat{k})(3 \hat{i}+2 \hat{j}+\hat{k})=\sqrt{1^2+1^2+(-1)^2} \sqrt{3^2+2^2+1^2} \cos \theta$

$\Rightarrow 3+2-1=\sqrt{3} \sqrt{9+4+1} \cos \theta$

$= \frac{4}{\sqrt{3} \sqrt{14}} =  \cos \theta$

$\cos \theta =\frac{4}{\sqrt{42}} \frac{\sqrt{42}}{\sqrt{42}} =\frac{2\sqrt{42}}{21}$

$\theta =\cos^{-1} \frac{2\sqrt{42}}{21}$