Let the direction cosines of line of intersection of the planes \(x-y=0\) and \(2 x+y+z=0\) are \(a_1, b_1, c_1\) so
\(a_1-b_1=0\)
and \(2 a_1+b_1+c_1=0\)
By cross-multiplication method, we have
\(\frac{a_1}{-1-0}=\frac{-b_1}{1-0}=\frac{c_1}{1+2} \Rightarrow \frac{a_1}{1}=\frac{b_1}{1}=\frac{c_1}{-3}\)
Similarly, let the direction cosines of line of intersection of the planes \(2 x-z=0\) and
\(\begin{aligned}
x+y-3 z & =0 \text { are } a_2, b_2, c_2 \text { so, } \\
2 a_2-b_2 & =0 \\
a_2+b_2-3 c_2 & =0
\end{aligned}\)
By cross-multiplication method, we have
\(\frac{a_2}{3-0}=\frac{-b_2}{-6-0}=\frac{c_2}{2+1} \Rightarrow \frac{a_2}{1}=\frac{b_2}{2}=\frac{c_2}{1}\)
So angle ' \(\theta\) ' between lines having directions cosines \(a_1, b_1, c_1\) and \(a_2, b_2, c_2\) respectively. is
\(\begin{aligned}
\theta & =\cos ^{-1}\left(\frac{\left|a_1 a_2+b_1 b_2+c_1 c_2\right|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right) \\
& =\cos ^{-1}\left(\frac{1+2-3}{\sqrt{1+1+9} \sqrt{1+4+1}}\right) \\
& =\cos ^{-1}(0)=90^{\circ}
\end{aligned}\)
Hence, option (d) is correct.