Let S be the reflection of a point Q with respect to the plane given by

$\overrightarrow{r}=(p-t)\hat{i}+t\hat{j}+(1-p)\hat{k}$

where t, p are real parameters and $\hat{i},\hat{j},\hat{k}$ are the unit vectors along the three positive coordinate axes. If the position vectors of Q and S are $10\hat{i}+15\hat{j}+20\hat{k}$ and $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ respectively, then which of the following is/are TRUE ?

To find which statements are true, we need to determine the Cartesian equation of the given plane and then compute the reflection of point Q with respect to this plane.

Step 1: Find the Cartesian equation of the plane.

The vector equation of the plane is given by:

$\overrightarrow{r}=(p-t)\hat{i}+t\hat{j}+(1-p)\hat{k}$

Comparing components of position vector $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$:

$x=p-t$

$y=t$

$z=1-p$

Adding the three parametric equations eliminates parameters $p$ and $t$:

$x+y+z=(p-t)+t+(1-p)=1$

Thus, the equation of the plane is:

$x+y+z-1=0$

Step 2: Find the reflection of point Q in the plane.

The position vector of point Q is $10\hat{i}+15\hat{j}+20\hat{k}$, so its coordinates are $Q(10,15,20)$.
Let $S(\alpha ,\beta ,\gamma )$ be the reflection of point Q in the plane $Ax+By+Cz+D=0$, where $A=1$, $B=1$, $C=1$, and $D=-1$.

Using the reflection formula for a point in a plane:

$\frac{\alpha -x_1}{A}=\frac{\beta -y_1}{B}=\frac{\gamma -z_1}{C}=\frac{-2(A{x}_1+B{y}_1+C{z}_1+D)}{A^2+B^2+C^2}$

Substitute the values of $Q(10,15,20)$ and the plane parameters into the formula:

$\frac{\alpha -10}{1}=\frac{\beta -15}{1}=\frac{\gamma -20}{1}=\frac{-2(10+15+20-1)}{1^2+1^2+1^2}$

$\frac{\alpha -10}{1}=\frac{\beta -15}{1}=\frac{\gamma -20}{1}=\frac{-2\left(44\right)}{3}=-\frac{88}{3}$

Now, solve for $\alpha$, $\beta$, and $\gamma$:

$\alpha =10-\frac{88}{3}=-\frac{58}{3}$

$\beta =15-\frac{88}{3}=-\frac{43}{3}$

$\gamma =20-\frac{88}{3}=-\frac{28}{3}$

Step 3: Evaluate the options.

1. Check $3(\alpha +\beta )$:

$3(\alpha +\beta )=3\left(-\frac{58}{3}-\frac{43}{3}\right)=-58-43=-101$

Thus, $3(\alpha +\beta )=-101$ is TRUE.

2. Check $3(\beta +\gamma )$:

$3(\beta +\gamma )=3\left(-\frac{43}{3}-\frac{28}{3}\right)=-43-28=-71$

Thus, $3(\beta +\gamma )=-71$ is TRUE.

3. Check $3(\gamma +\alpha )$:

$3(\gamma +\alpha )=3\left(-\frac{28}{3}-\frac{58}{3}\right)=-28-58=-86$

Thus, $3(\gamma +\alpha )=-86$ is TRUE.

4. Check $3(\alpha +\beta +\gamma )$:

$3(\alpha +\beta +\gamma )=3\left(-\frac{58}{3}-\frac{43}{3}-\frac{28}{3}\right)=-129\ne -121$

Thus, $3(\alpha +\beta +\gamma )=-121$ is FALSE.

Therefore, the correct options are:

$3(\alpha +\beta )=-101$
$3(\beta +\gamma )=-71$
$3(\gamma +\alpha )=-86$