Explanation
To find the equation of a line, we need a point on the line and its direction vector.
Step 1: Identify the point and direction vector
The line passes through the point (x1,y1,z1)=(−1,2,3).
Since the line is perpendicular to the plane 2x+3y+z+5=0, the direction vector of the line (v) is the same as the normal vector to the plane (n).
The normal vector to the plane ax+by+cz+d=0 is n=ai^+bj^+ck^.
From the given plane equation, the normal vector is n=2i^+3j^+1k^. Thus, the direction ratios of the line are (2,3,1).
Step 2: Write the equation of the line
The symmetric form of a line passing through (x1,y1,z1) with direction ratios (a,b,c) is:
ax−x1=by−y1=cz−z1
Substituting our values:
2x−(−1)=3y−2=1z−3
2x+1=3y−2=1z−3
Step 3: Compare with the given options
None of the options exactly match this form, so we manipulate our derived equation:
Let 2x+1=3y−2=1z−3=k.
Multiplying by 6 (the least common multiple of 2,3,1):
3(x+1)=2(y−2)=6(z−3)
3x+3=2y−4=6z−18
