Explanation
Any plane through $$\left( 1,0,0 \right) $$ is
$$A\left( x-1 \right) +B\left( y-0 \right) +C\left( z-0 \right) =0$$ ...$$(1)$$
It contains $$\left( 0,1,0 \right) $$ if $$-A+B=0$$ ....$$(2)$$
Also, $$(1)$$ makes an angle of $$\displaystyle \dfrac { \pi }{ 4 } $$ with the plane $$x+y=3$$
Therefore, $$\displaystyle \cos { \dfrac { \pi }{ 4 } } =\dfrac { \left| A+B \right| }{ \sqrt { { A }^{ 2 }+{ B }^{ 2 }+{ C }^{ 2 } } \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } } $$
$$\Rightarrow { \left( A+B \right) }^{ 2 }={ A }^{ 2 }+{ B }^{ 2 }+{ C }^{ 2 }\Rightarrow 2AB={ C }^{ 2 }$$ ....$$(3)$$
From (2) and (3), $${ C }^{ 2 }=2{ A }^{ 2 }\Rightarrow C=\pm \sqrt { 2 } A$$
Hence, $$A:B:C::A:A:\pm \sqrt { 2 } A$$
$$\therefore$$ direction ratios are $$1:1:\pm \sqrt { 2 } $$
Eliminating $$l$$ from equations $$n=l+m, \; m=2l+3n$$, we get
$$\dfrac{m}{n}= \dfrac{5}{3}$$...$$(1)$$
Similarly, eliminating $$m$$ from the two equations we get
$$ \dfrac{n}{l}=\dfrac{3}{-2}$$...$$(2)$$
From $$(1)$$ and $$(2)$$, $$m:n:l = 5:3:-2$$
Let $$m=5t,\; n=3t,\; l=-2t$$
Now, use $$l^2+m^2+n^2=1$$
$$ \Rightarrow (25+9+4)t^2=1$$
$$ \Rightarrow t^2= \dfrac{1}{38}$$
$$ \Rightarrow t= \pm \dfrac{1}{\sqrt{38}}$$
Positive value of $$t$$ will give direction cosines of one line and negative value will give direction cosines of another line.
Hence, they will be parallel to each other with angle between them $$180^{o}$$
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