Search any question & find its solution
Question:
Answered & Verified by Expert
If the point $P(\alpha, \beta, \gamma)$ lies on the plane $2 x+y+z=1$ and $[\alpha \beta \gamma]\left[\begin{array}{lll}1 & 9 & 1 \\ 8 & 2 & 1 \\ 7 & 3 & 1\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$, then $\alpha^2+\beta^2+\gamma^2=$
Options:
Solution:
2892 Upvotes
Verified Answer
The correct answer is:
86
Point $P(\alpha, \beta, \gamma)$ lies on plane $2 x+y+z=1 \ldots$
Form Eqs. (i) and (iv), we get
$$
2 \alpha+\beta+\gamma=1 \quad \Rightarrow \alpha+\beta+\gamma=0 \quad \Rightarrow \alpha=1
$$
put in Eqs. (ii) and (iii), we get
$$
1+8 \beta+7 \gamma=0
$$
and $\quad 9(1)+2 \beta+3 \gamma=0$
$$
\begin{aligned}
\text { Eq (V) }-4 \times \text { Eq. (vii) } & \\
8 \beta+7 \gamma & =-1 \\
8 \beta+12 \gamma & =-36 \\
& =-1 \\
\hline-5 \gamma & =35 \\
\Rightarrow \quad \gamma & =-7
\end{aligned}
$$
Put value of $\gamma$ in Eqs. (vi), we get
$$
2 \beta+3(-7)=-9 \Rightarrow 2 \beta=12 \Rightarrow \beta=6
$$
Now, $\alpha^2+\beta^2+\gamma^2$
$$
=(1)^2+(6)^2+(-7)^2=1+36+49=86
$$
Form Eqs. (i) and (iv), we get
$$
2 \alpha+\beta+\gamma=1 \quad \Rightarrow \alpha+\beta+\gamma=0 \quad \Rightarrow \alpha=1
$$
put in Eqs. (ii) and (iii), we get
$$
1+8 \beta+7 \gamma=0
$$
and $\quad 9(1)+2 \beta+3 \gamma=0$
$$
\begin{aligned}
\text { Eq (V) }-4 \times \text { Eq. (vii) } & \\
8 \beta+7 \gamma & =-1 \\
8 \beta+12 \gamma & =-36 \\
& =-1 \\
\hline-5 \gamma & =35 \\
\Rightarrow \quad \gamma & =-7
\end{aligned}
$$
Put value of $\gamma$ in Eqs. (vi), we get
$$
2 \beta+3(-7)=-9 \Rightarrow 2 \beta=12 \Rightarrow \beta=6
$$
Now, $\alpha^2+\beta^2+\gamma^2$
$$
=(1)^2+(6)^2+(-7)^2=1+36+49=86
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.