Question

Let$L$ be a line obtained from the intersection of two planes$x+2y+z=6$ and$y+2z=4.$ If point $P(ɑ,\beta ,\gamma )$is the foot of the perpendicular from$(3,2,1)$ on $L$, then the value of$21(ɑ+\beta +\gamma )$ equal:

• A. $142$
• B. $68$
• C. $136$
• D. $102$

Answer 1

The correct option is D

$102$

Explanation for the correct option:

Finding the value of $21(ɑ+\beta +\gamma )$:

Given equation of the line is $x+2y+z–6=0=y+2z-4$

$\frac{[x+2]}{3}=\frac{[y–4]}{-2}=\frac{z}{1}=\lambda$

Directional ration of $PQ:3\lambda –5,–2\lambda –2,\lambda –1$

Directional ratio of y line are $(3,-2,1)$
$$\begin{gathered} x+2y+z=6 \\ (y+2z=4)\times 2 \\ x-3z=-2 \\ \Rightarrow x=3z-2 \\ \Rightarrow y=4-2z \\ 3x+2=z-2y-4=z \end{gathered}$$

Line of intersection of two planes is
$3x+2=-2y-4=z=\lambda \left[Let\right]$
$\because AP\perp \text{line}$
$$\begin{gathered} \therefore AP\cdot (3\hat{i}-2{\hat{j}}^{}+\hat{k})=0 \\ (3\lambda -5)\times 3+(-2\lambda +2)(-2)+(\lambda -1)\times 1=0 \\ 9\lambda -15+4\lambda -4+\lambda -1=0 \\ 14\lambda =20 \\ \lambda =\frac{10}{7} \\ \Rightarrow P\left(\frac{16}{7},\frac{8}{7},\frac{10}{7}\right) \\ \Rightarrow \alpha +\beta +\gamma =\frac{16+8+10}{7}=\frac{34}{7} \\ 21(\alpha +\beta +\gamma )=102 \end{gathered}$$

Therefore, the correct answer is option (D).