Question:

Step 1: Determine the equation of the family of planes passing through the intersection of the given planes. The equation of a plane passing through the intersection of $x + y + z = 1$ (or $x + y + z - 1 = 0$) and $2x + 3y - z + 4 = 0$ is given by: $$(x + y + z - 1) + \lambda(2x + 3y - z + 4) = 0$$ Rearranging the terms, we get: $$(1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (4\lambda - 1) = 0$$ Step 2: Apply the condition that the plane is parallel to the Y-axis. A plane is parallel to the Y-axis if its normal vector is perpendicular to the direction vector of the Y-axis, which is $\mathbf{j} = (0, 1, 0)$. This implies that the coefficient of $y$ in the plane's equation must be zero. Therefore, we set the coefficient of $y$ to zero: $$1 + 3\lambda = 0$$ $$\lambda = -\frac{1}{3}$$ Step 3: Substitute the value of $\lambda$ to find the equation of the required plane. Substitute $\lambda = -\frac{1}{3}$ into the plane equation from Step 1: $$\left(1 + 2\left(-\frac{1}{3}\right)\right)x + \left(1 + 3\left(-\frac{1}{3}\right)\right)y + \left(1 - \left(-\frac{1}{3}\right)\right)z + \left(4\left(-\frac{1}{3}\right) - 1\right) = 0$$ $$\left(1 - \frac{2}{3}\right)x + (1 - 1)y + \left(1 + \frac{1}{3}\right)z + \left(-\frac{4}{3} - 1\right) = 0$$ $$\left(\frac{1}{3}\right)x + (0)y + \left(\frac{4}{3}\right)z + \left(-\frac{7}{3}\right) = 0$$ Multiplying the entire equation by 3 to clear the denominators, we obtain the equation of the plane: $$x + 4z - 7 = 0$$ Step 4: Verify if the point $(-3, 0, -1)$ lies on the plane. Substitute the coordinates of the point $(-3, 0, -1)$ into the plane equation $x + 4z - 7 = 0$: $$(-3) + 4(-1) - 7$$ $$-3 - 4 - 7$$ $$-14$$ Since $-14 \neq 0$, the point $(-3, 0, -1)$ does not satisfy the equation of the plane.