Step 1: Find the Cartesian equation of the plane.

The vector equation of the plane is given as: $\vec{r}=-(t+p) \hat{\imath}+t \hat{\jmath}+(1+p) \hat{k}$ We can rewrite this equation by separating the terms with parameters $t$ and $p$: $\vec{r} = (-t \hat{\imath} + t \hat{\jmath}) + (-p \hat{\imath} + p \hat{k}) + 1 \hat{k}$ $\vec{r} = (0 \hat{\imath} + 0 \hat{\jmath} + 1 \hat{k}) + t(-\hat{\imath} + \hat{\jmath} + 0 \hat{k}) + p(-\hat{\imath} + 0 \hat{\jmath} + 1 \hat{k})$ This is the parametric form of the plane, $\vec{r} = \vec{a} + t\vec{b} + p\vec{c}$, where:

• $\vec{a} = \hat{k}$ is the position vector of a point on the plane, so the point is $A(0, 0, 1)$.
• $\vec{b} = -\hat{\imath} + \hat{\jmath}$ and $\vec{c} = -\hat{\imath} + \hat{k}$ are two vectors parallel to the plane.

A normal vector to the plane, $\vec{n}$, can be found by taking the cross product of $\vec{b}$ and $\vec{c}$: $\vec{n} = \vec{b} \times \vec{c} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$ $\vec{n} = \hat{\imath}(1-0) - \hat{\jmath}(-1-0) + \hat{k}(0 - (-1)) = \hat{\imath} + \hat{\jmath} + \hat{k}$ The Cartesian equation of a plane passing through a point $(x_0, y_0, z_0)$ with a normal vector $(a, b, c)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$. Using point $A(0, 0, 1)$ and normal vector $\vec{n}=(1, 1, 1)$: $1(x-0) + 1(y-0) + 1(z-1) = 0$ $x + y + z - 1 = 0$So, the equation of the plane is $x + y + z = 1$.

Step 2: Find the coordinates of the reflection point S.

Let the point $Q$ be $(x_1, y_1, z_1) = (10, 15, 20)$. Let its reflection $S$ be $(\alpha, \beta, \gamma)$. The formula for the reflection of a point $(x_1, y_1, z_1)$ in the plane $ax+by+cz+d=0$ to a point $(x_2, y_2, z_2)$ is: $\frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = \frac{z_2 - z_1}{c} = -2\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$ Here, the plane is $x+y+z-1=0$, so $a=1, b=1, c=1, d=-1$. The point $Q$ is $(10, 15, 20)$. The reflection $S$ is $(\alpha, \beta, \gamma)$.

Substituting these values into the formula: $\frac{\alpha - 10}{1} = \frac{\beta - 15}{1} = \frac{\gamma - 20}{1} = -2\frac{1(10)+1(15)+1(20)-1}{1^2+1^2+1^2}$ $\frac{\alpha - 10}{1} = \frac{\beta - 15}{1} = \frac{\gamma - 20}{1} = -2\frac{10+15+20-1}{3}$ $\frac{\alpha - 10}{1} = \frac{\beta - 15}{1} = \frac{\gamma - 20}{1} = -2\frac{44}{3} = -\frac{88}{3}$

Now, we can find $\alpha, \beta,$ and $\gamma$:

• $\alpha - 10 = -\frac{88}{3} \implies \alpha = 10 - \frac{88}{3} = \frac{30-88}{3} = -\frac{58}{3}$
• $\beta - 15 = -\frac{88}{3} \implies \beta = 15 - \frac{88}{3} = \frac{45-88}{3} = -\frac{43}{3}$
• $\gamma - 20 = -\frac{88}{3} \implies \gamma = 20 - \frac{88}{3} = \frac{60-88}{3} = -\frac{28}{3}$ So, the position vector of $S$ is $-\frac{58}{3} \hat{\imath} - \frac{43}{3} \hat{\jmath} - \frac{28}{3} \hat{k}$.

Step 3: Verify the given options.

A: $3(\alpha+\beta)=-101$ $3(\alpha+\beta) = 3\left(-\frac{58}{3} - \frac{43}{3}\right) = 3\left(-\frac{101}{3}\right) = -101$ This statement is TRUE.

B: $3(\beta+\gamma)=-71$ $3(\beta+\gamma) = 3\left(-\frac{43}{3} - \frac{28}{3}\right) = 3\left(-\frac{71}{3}\right) = -71$ This statement is TRUE.

C: $3(\gamma+\alpha)=-86$ $3(\gamma+\alpha) = 3\left(-\frac{28}{3} - \frac{58}{3}\right) = 3\left(-\frac{86}{3}\right) = -86$ This statement is TRUE.

D: $3(\alpha+\beta+\gamma)=-121$ $3(\alpha+\beta+\gamma) = 3\left(-\frac{58}{3} - \frac{43}{3} - \frac{28}{3}\right) = 3\left(-\frac{129}{3}\right) = -129$ Since $-129 \neq -121$, this statement is FALSE.

Therefore, the correct options are A, B, and C.