Equation of line passing through the line x + y + z + 3=0 and 2x–y + 3z + 1=0 is given by,

(x + y + z + 3) + k(2x–y + 3z + 1)=0 …………………….(1)

x(1 + 2k) + y(1 - k) + z(1 + 3k) + 3 + k=0 [k is a constant]

Again, the required plane is parallel to the line

So, we should have,

[1×(1 + 2k)] + [2×(1 - k)] + [3×(1 + 3k)]=0

1 + 2k + 2 - 2k + 3 + 9k=0

9k= - 6

Putting in equation (1) we get,

3(x + y + z + 3) - 2(2x–y + 3z + 1)=0

3x + 3y + 3z + 9 - 4x + 2y - 6z - 2=0

- x + 5y - 3z + 7=0

x - 5y + 3z - 7=0

x - 5y + 3z=7

∴The equation of the plane through the line x + y + z + 3 = 0 = 2x – y + 3z + 1 and parallel to the line

is x - 5y + 3z=7.