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!

$$
\frac {X+1}4 = \frac {Y-2}{-2} = \frac {Z+6}7 = k \\
X = 4k - 1 \\
Y = -2k + 2 \\
Z = 7k - 6
$$
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 $(3,0,1)$.
