# Question #b1688

Apr 27, 2015

The answer is: $\left(3 , - 2 , 4\right)$.

First of all let's write the line in parametric form.

If we put $x = t$, than:

$\frac{t}{3} = \frac{y + 7}{5} \Rightarrow y = - 7 + \frac{5}{3} t$

and:

$\frac{t}{3} = \frac{z - 2}{2} \Rightarrow z = 2 + \frac{2}{3} t$.

So the line is:

$x = t$
$y = - 7 + \frac{5}{3} t$
$z = 2 + \frac{2}{3} t$

the direction of this line is $\left(1 , \frac{5}{3} , \frac{2}{3}\right)$.

To write a plane, given a point $P \left({x}_{p} , {y}_{p} , {z}_{p}\right)$ and the direction $\vec{v} \left(a , b , c\right)$ of a perpendicular at the plane itself, we can use this formula:

$a \left(x - {x}_{p}\right) + b \left(y - {y}_{p}\right) + c \left(z - {z}_{p}\right) = 0$,

so:

$1 \left(x - 2\right) + \frac{5}{3} \left(y + 1\right) + \frac{2}{3} \left(z - 3\right) = 0 \Rightarrow$

$3 x - 6 + 5 y + 5 + 2 z - 6 = 0 \Rightarrow 3 x + 5 y + 2 z + 7 = 0$.

Now, to find the point, it is necessary to make a system with the plane and the given line:

$3 x + 5 y + 2 z + 7 = 0$
$x = t$
$y = - 7 + \frac{5}{3} t$
$z = 2 + \frac{2}{3} t$

$\Rightarrow 3 t - 35 + \frac{25}{3} t + 4 + \frac{4}{3} t - 7 = 0 \Rightarrow \frac{38}{3} t = 38 \Rightarrow t = 3$.

Now let's substitue this value of $t$ in the three equation of the line:

$x = 3$
$y = - 7 + \frac{5}{3} \cdot 3 = - 2$
$z = 2 + \frac{2}{3} \cdot 3 = 4$.

The point is: $\left(3 , - 2 , 4\right)$.