Answer:
The point of intersection [tex]P\left(\dfrac{1133}{122},\dfrac{149}{122},\dfrac{699}{122}\right)[/tex]
Step-by-step explanation:
Equation of line:
[tex]x=7+9t[/tex]
[tex]y=3-7t[/tex]
[tex]z=7-5t[/tex]
Equation of plane:
[tex]5x-6y-7z=-1[/tex]
We need to find the point of intersection of line and plane.
Point of intersection: When both line and plane meet at single point.
So, put the value of x, y and z into plane.
[tex]5(7+9t)-6(3-7t)-7(7-5t)=-1[/tex]
[tex]35+45t-18+42t-49+35t=-1[/tex]
[tex]122t=-1+32[/tex]
[tex]t=\dfrac{31}{122}[/tex]
Substitute the value of t into x, y and z
[tex]x=7+9\cdot \dfrac{31}{122}=\dfrac{1133}{122}[/tex]
[tex]y=3-7\cdot \dfrac{31}{122}=\dfrac{149}{122}[/tex]
[tex]z=7-5\cdot \dfrac{31}{122}=\dfrac{699}{122}[/tex]
Point of intersection:
[tex]\left(\dfrac{1133}{122},\dfrac{149}{122},\dfrac{699}{122}\right)[/tex]
Hence, The point of intersection [tex]P\left(\dfrac{1133}{122},\dfrac{149}{122},\dfrac{699}{122}\right)[/tex]
