Question

$L:\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+1}{4}$
${\pi }_1:x+2y+3z=14,{\pi }_2:2x-y+3z=27$

If the line $L$ meets the plane ${\pi }_1$ in the point $P$, and the coordinates of $P$ are $(\alpha ,\beta ,\gamma )$, then ${\alpha }^2+{\beta }^2+{\gamma }^2$ is equal to

• A. $3$
• B. $14$
• C. $28$
• D. $29$

Answer 1

Answer: (B)

$14$

$L=\frac{x-1}{2}=\frac{y+1}{3}=\frac{z+1}{4}$ ........(i) and plane

$\pi \Rightarrow x+2y+3z=14$ .......(ii)

Given that $L$ meets plane ${\pi }_1$ (means intersection points) so from $e{q}^n$ (i)

$\Rightarrow \frac{x+1}{2}=\frac{y+1}{3}=\frac{x+1}{4}=K$

$\Rightarrow x=2k-1andy=3k-1,z=4k-1$

putting this values in ..... (ii)

So $(2k-1)+2(3k-1)+3(4k-1)=14$

$\Rightarrow 2k-1+6k-2+12k-3=14$

$20k-6=14$

$\Rightarrow 20k=20\to k=1$

so points $x=1,y=2,z=3$ inform of $\alpha ,\beta ,\gamma =\alpha =1,\beta =2,\gamma =3$

so ${\alpha }^2+{\beta }^2+{\gamma }^2=1^2+2^2+3^2$

$=14$