Find the coordinates of the point where the line through

It is known that the equation of the line through the points, $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$, is $\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$

Since the line passes through the points, (3, −4, −5) and (2, −3, 1), its equation is given by,

$\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}$

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

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

Therefore, 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

∴ 2 (3 − k) + (k − 4) + (6k − 5) = 7

$\Rightarrow 5 k-3=7$

$\Rightarrow k=2$

Hence, the coordinates of the required point are (3 − 2, 2 − 4, 6 × 2 − 5) i.e.,

(1, −2, 7).