If l, m, n are the direction cosines of a lin | vedclass

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(B) Given that $(l, m, n)$ are the direction cosines of a line perpendicular to two lines with direction ratios $(1, 2, -1)$ and $(1, -2, 1)$.Since th

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$\frac{9}{5}$

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(B) Given that $(l, m, n)$ are the direction cosines of a line perpendicular to two lines with direction ratios $(1, 2, -1)$ and $(1, -2, 1)$.Since the line is perpendicular to both,we have:$l + 2m - n = 0$  ...$(i)$$l - 2m + n = 0$  ...$(ii)$Adding $(i)$ and $(ii)$,we get $2l = 0$,which implies $l = 0$.Substituting $l = 0$ in $(i)$,we get $2m - n = 0$,so $n = 2m$.We know that for direction cosines,$l^2 + m^2 + n^2 = 1$.Substituting $l = 0$ and $n = 2m$,we get $0^2 + m^2 + (2m)^2 = 1$.$m^2 + 4m^2 = 1 \Rightarrow 5m^2 = 1 \Rightarrow m^2 = \frac{1}{5}$.Now,we need to find $(l + m + n)^2$.$(l + m + n)^2 = (0 + m + 2m)^2 = (3m)^2 = 9m^2$.Substituting $m^2 = \frac{1}{5}$,we get $(l + m + n)^2 = 9 	imes \frac{1}{5} = \frac{9}{5}$.

If $l, m, n$ are the direction cosines of a line that is perpendicular to the lines having the direction ratios $1, 2, -1$ and $1, -2, 1$,then $(l+m+n)^2=$

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