Question:

Step 1: Identify the direction vectors of the given lines. The first line is given as $\dfrac{x}{3} = \dfrac{y+1}{0} = \dfrac{z-2}{-1}$, which has direction vector $\vec{d_1} = (3, 0, -1)$. The second line is given as $\dfrac{x-1}{1} = \dfrac{y+1}{2} = \dfrac{z+1}{-1}$, which has direction vector $\vec{d_2} = (1, 2, -1)$. Step 2: Find the normal vector to the plane using the cross product. Since the plane is parallel to both lines, its normal vector must be perpendicular to both direction vectors. We calculate: $$\vec{n} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -1 \\ 1 & 2 & -1 \end{vmatrix}$$ Expanding the determinant: $$\vec{n} = \hat{i}(0 \cdot (-1) - (-1) \cdot 2) - \hat{j}(3 \cdot (-1) - (-1) \cdot 1) + \hat{k}(3 \cdot 2 - 0 \cdot 1)$$ $$= \hat{i}(0 + 2) - \hat{j}(-3 + 1) + \hat{k}(6)$$ $$= 2\hat{i} + 2\hat{j} + 6\hat{k}$$ Simplifying by dividing by 2, we get $\vec{n} = (1, 1, 3)$. Step 3: Write the equation of the plane passing through the given point. The plane passes through the point $(-1, 2, 0)$ with normal vector $(1, 1, 3)$. Using the point-normal form: $$1(x - (-1)) + 1(y - 2) + 3(z - 0) = 0$$ $$x + 1 + y - 2 + 3z = 0$$ $$x + y + 3z = 1$$ Step 4: Identify coefficients and calculate the final answer. From the equation $x + y + 3z = 1$, we have: - $a = 1$ - $b = 1$ - $c = 3$ Therefore: $$a + b + c = 1 + 1 + 3 = 5$$ The value of $(a + b + c)$ is **5**, which corresponds to **Option 3**.