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A line with direction cosines proportional to $2,1,2$ meets each of the lines $x=y+a=z$ and $x+a=2 y=2 z$. The co-ordinates of each of the point of intersection are given by
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The correct answer is:
$(3 a, 2 a, 3 a),(a, a, a)$
$(3 a, 2 a, 3 a),(a, a, a)$
Any point on the line $\frac{x}{1}=\frac{y+a}{1}=\frac{z}{1}=t_1 \quad$ (say) is $\left(t_1, t_1-a, t_1\right)$ and any point on the line $\frac{x+a}{2}=\frac{y}{1}=\frac{z}{1}=t_2 \quad($ say $)$ is $\left(2 t_2-a, t_2, t_2\right)$.
Now direction cosine of the lines intersecting the above lines is proportional to $\left(2 t_2-a-t_1, t_2-t_1+a, t_2-t_1\right)$.
Hence $2 \mathrm{t}_2-\mathrm{a}-\mathrm{t}_1=2 \mathrm{k}, \mathrm{t}_2-\mathrm{t}_1+\mathrm{a}=\mathrm{k}$ and $\mathrm{t}_2-\mathrm{t}_1=2 \mathrm{k}$
On solving these, we get $t_1=3 a, t_2=a$.
Hence points are $(3 a, 2 a, 3 a)$ and $(a, a, a)$
Now direction cosine of the lines intersecting the above lines is proportional to $\left(2 t_2-a-t_1, t_2-t_1+a, t_2-t_1\right)$.
Hence $2 \mathrm{t}_2-\mathrm{a}-\mathrm{t}_1=2 \mathrm{k}, \mathrm{t}_2-\mathrm{t}_1+\mathrm{a}=\mathrm{k}$ and $\mathrm{t}_2-\mathrm{t}_1=2 \mathrm{k}$
On solving these, we get $t_1=3 a, t_2=a$.
Hence points are $(3 a, 2 a, 3 a)$ and $(a, a, a)$
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