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

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 $\;(3,0,1)\;\ldots$
