**Question:**

Find the coordinates of the point where the line through $(5,1,6)$ and

$(3,4,1)$ crosses the $Z X$ - plane.

**Solution:**

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

The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,

$\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}$

$\Rightarrow \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k$ (say)

$\Rightarrow x=5-2 k, y=3 k+1, z=6-5 k$

Any point on the line is of the form (5 − 2*k*, 3*k* + 1, 6 −5*k*).

Since the line passes through ZX-plane,

$3 k+1=0$

$\Rightarrow k=-\frac{1}{3}$

$\Rightarrow 5-2 k=5-2\left(-\frac{1}{3}\right)=\frac{17}{3}$

$6-5 k=6-5\left(-\frac{1}{3}\right)=\frac{23}{3}$

Therefore, the required point is $\left(\frac{17}{3}, 0, \frac{23}{3}\right)$.

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