The plane through intersection is \((x+y+z-1) + \lambda(2x+3y-z+4) = 0\), giving \((1+2\lambda)x + (1+3\lambda)y + (1-\lambda)z + (4\lambda-1) = 0\). For this to be parallel to y-axis, the coefficient of y must be zero: \(1+3\lambda = 0\), so \(\lambda = -\frac{1}{3}\). The plane equation becomes \((1-\frac{2}{3})x + 0 \cdot y + (1+\frac{1}{3})z + (-\frac{4}{3}-1) = 0\), which is \(\frac{1}{3}x + \frac{4}{3}z - \frac{7}{3} = 0\), or \(x + 4z = 7\). Checking option A: \(-3 + 4(-1) = -7 \neq 7\). Hmm, let me recalculate. The plane is \(x + 4z - 7 = 0\). For \((-3,0,-1)\): \(-3-4 = -7\), which gives \(-7 \neq 7\). Let me check the calculation of \(\lambda\). Actually, I think I made a sign error. Rechecking: plane is \(x+4z=7\) or possibly different. Testing options directly in the family of planes would be more reliable.