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12. Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.

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We know that the equation of the line that passes through the points $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ is given by the relation;

$\frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}$

and the line passing through the points, $(3,-4,-5)\ and\ (2,-3,1)$ .

$\implies \frac{x-3}{2-3} = \frac{y+4}{-3+4} = \frac{z+5}{1+5} = k\ (say)$

$\implies \frac{x-3}{-1} = \frac{y+4}{-1} = \frac{z+5}{6} = k\ (say)$

$\implies x=3-k,\ y=k-4,\ z=6k-5$

And any point on the line is of the form. $(3-k,k-4,6k-5)$

This point lies on the plane, $2x+y+z = 7$

$\therefore 2(3-k)+(k-4)+(6k-5) = 7$

$\implies 5k-3=7$

or $k =2$ .

Hence, the coordinates of the required point are $(3-2,2-4,6(2)-5)$ or $(1,-2,7)$ .

Posted by

Divya Prakash Singh

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