Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The direction ratios of the line passing through $A(1,p,1)$ and $B(2,5,-1)$ are given by $(2-1,5-p,-1-1)=(1,5-p,-2)$.

Since the line is perpendicular to the vector $4\hat{i}-2\hat{j}+\hat{k}$, the dot product of their direction vectors is zero:

$1(4)+(5-p)(-2)+(-2)(1)=0$

$4-10+2p-2=0$

$2p=8\Rightarrow p=4$

Substituting $p=4$, the direction ratios of the line are $(1,1,-2)$.

Since the points $A,B,C$ are collinear, the direction ratios of the line segment $BC$ must be proportional to $(1,1,-2)$. The direction ratios of $BC$ are $(q-2,11-5,r-(-1))=(q-2,6,r+1)$.

Equating the ratios:

$\frac{q-2}{1}=\frac{6}{1}=\frac{r+1}{-2}$

From $\frac{q-2}{1}=6$, we get $q=8$.

From $\frac{r+1}{-2}=6$, we get $r+1=-12\Rightarrow r=-13$.

Therefore, the value of $p+q+r$ is $4+8-13=-1$.

Answer: $-1$