Line l is r=-8i+5j-6k+t(5i-2k)
Plane is -x-2y+5z = 0
What is the point of intersection? I get (-56/3, 5, -26/15) but my student's school teacher has a different answer of (-23,5,0)
Who is right?
The line is given by x= -8+5t, y= 5, z= -6- 2t.
Putting those into the equation of the plane, -(-8+ 5t)- 2(5)+ 5(-6- 2t)= 8- 5t- 10- 30- 10t= -15t- 32= 0 so that t= -32/15 and then
x= -8+ 5(-32/15)= -8- 32/3= (-24- 32)/3= -56/3.
y= 5
z= -6- 2(-32/15)= (-90+ 64)/15= -26/15
Your line is
$\displaystyle \displaystyle \begin{align*} \begin{cases} x &= -8 + 5t \\ y &= \phantom{-} 5 \\ z &= -6 -2t \end{cases} \end{align*}$
Substituting into the plane we find
$\displaystyle \displaystyle \begin{align*} -x - 2y + 5z &= 0 \\ - \left( -8 + 5t \right) - 2 \left( 5 \right) + 5 \left( -6 - 2t \right) &= 0 \\ 8 - 5t - 10 - 30 - 10t &= 0 \\ -32 - 15t &= 0 \\ -15t &= 32 \\ t &= -\frac{32}{15} \end{align*}$
So that gives the point of intersection as
$\displaystyle \displaystyle \begin{align*} \begin{cases} x &= -8 + 5 \left( -\frac{32}{15} \right) \\ y &= \phantom{-} 5 \\ z &= -6 - 2 \left( -\frac{32}{15} \right) \end{cases} \\ \begin{cases} x &= -\frac{24}{3} - \frac{32}{3} \\ y &= \phantom{-} 5 \\ z &= -\frac{90}{15} + \frac{64}{15} \end{cases} \\ \begin{cases} x &= -\frac{ 56}{3} \\ y &= \phantom{-} 5 \\ z &= -\frac{26}{15} \end{cases} \end{align*}$
So the point of intersection is $\displaystyle \displaystyle \begin{align*} \left( x, y, z \right) = \left( -\frac{56}{3} , 5, -\frac{26}{15} \right) \end{align*}$. I agree with your answer