The equation of the plane passing through the intersection of the planes $x+y+z=6$ and $2x + 3y + 4z + 5 = 0$ and the point $(1,1,1)$ is:

Solution

1. The equation of any plane passing through the intersection of $x+y+z-6=0$ and $2x + 3y + 4z + 5 = 0$ is given by $(x+y+z-6)+\lambda (2x+3y+4z+5)=0$.
2. Substitute the point $(1,1,1)$ into the equation to find $\lambda$: $(1+1+1-6)+\lambda (2(1)+3(1)+4(1)+5)=0$, which simplifies to $-3+14\lambda =0$, so $\lambda =\frac{3}{14}$.
3. Substitute $\lambda =\frac{3}{14}$ back into the family equation: $(x+y+z-6)+\frac{3}{14}(2x+3y+4z+5)=0$.
4. Multiply the entire equation by 14 to clear the fraction: $14(x + y + z - 6) + 3(2x + 3y + 4z + 5) = 0$, resulting in $14x + 14y + 14z - 84 + 6x + 9y + 12z + 15 = 0$.
5. Combine like terms to obtain $20x + 23y + 26z - 69 = 0$.
6. Hence the answer is (A).