Find the point of intersection of the straight line $\frac{X+1}{4}=\frac{Y-2}{-2}=\frac{Z+6}{7}$ and plane $3X+8Y-9Z=0$

Question

Find the point of intersection of the straight line $$\frac{X+1}{4}=\frac{Y-2}{-2}=\frac{Z+6}{7}$$ and plane $3X+8Y-9Z=0$
the point of the line is $M(-1,2,-6)$ and direction vector of the line is $A(4,-2,7)$
I would like to get some advice how to do that.
Thanks!

Answer 1

$$ \frac {X+1}4 = \frac {Y-2}{-2} = \frac {Z+6}7 = k \\ X = 4k - 1 \\ Y = -2k + 2 \\ Z = 7k - 6 $$ Substitute it to the equation of plane $$ 3X+8Y-9Z = 0 \\ 12k-3 - 16k + 16 - 63k + 54 = 0 \\ -67k + 67 = 0 \\ k = 1 $$ Since you know $k$, you can easily find $X,Y,Z$. $$ X = 4k-1 = 3 \\ Y = -2k+2 = 0 \\ Z = 7k-6 = 1 $$ So, point of intersection is $(3,0,1)$.

Answer 2

Write all the parameters as one of them, say:

$$y=\frac{-x-1}2+2=\frac{-x+3}2\;,\;\;z=\frac{7x+7}4-6=\frac{7x-17}{4}$$

Substitute in the plane's formula:

$$3x+8\left(\frac{-x+3}2\right)-9\left(\frac{7x-17}4\right)=0\iff -\frac{67}4x+\frac{201}4=0\implies$$

$$ x=3\;,\;y=0\;,\;z=1$$

From here it follows that I've no idea what you call $\,M\,$ to as the intersection point between the given line and given plane is $\;(3,0,1)\;\ldots$