Let $$\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$$ and $$\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$$. Let a vector $$\vec{v}$$ be in the plane containing $$\vec{a}$$ and $$\vec{b}$$. If $$\vec{v}$$ is perpendicular to the vector $$3\hat{i} + 2\hat{j} - \hat{k}$$ and its projection on $$\vec{a}$$ is 19 units, then $$|2\vec{v}|^2$$ is equal to _________.

We have the two fixed vectors $$\vec a=2\hat i-\hat j+2\hat k$$ and $$\vec b=\hat i+2\hat j-\hat k$$. Because the required vector $$\vec v$$ lies in the plane containing $$\vec a$$ and $$\vec b$$, every such vector can be written as a linear combination of these two. So we introduce real numbers $$p$$ and $$q$$ and write

$$\vec v=p\,\vec a+q\,\vec b.$$

The statement “$$\vec v$$ is perpendicular to $$3\hat i+2\hat j-\hat k$$” means that the dot-product of these two vectors is zero. Using the bilinearity of the dot product we have

$$\vec v\cdot\left(3\hat i+2\hat j-\hat k\right)=\left(p\vec a+q\vec b\right)\cdot\left(3\hat i+2\hat j-\hat k\right) = p\bigl(\vec a\cdot(3\hat i+2\hat j-\hat k)\bigr)+q\bigl(\vec b\cdot(3\hat i+2\hat j-\hat k)\bigr)=0.$$

Let us evaluate each scalar product appearing above. First,

$$\vec a\cdot(3\hat i+2\hat j-\hat k)=2\cdot3+(-1)\cdot2+2\cdot(-1)=6-2-2=2,$$

and secondly,

$$\vec b\cdot(3\hat i+2\hat j-\hat k)=1\cdot3+2\cdot2+(-1)\cdot(-1)=3+4+1=8.$$

Substituting these values we get

$$2p+8q=0,$$

which simplifies to

$$p+4q=0\quad\Longrightarrow\quad p=-4q.$$

The next piece of information is that the projection of $$\vec v$$ on $$\vec a$$ has magnitude $$19$$. The length of the projection of a vector $$\vec u$$ on another vector $$\vec w$$ is given by the formula

$$\text{projection length}=\frac{|\vec u\cdot\vec w|}{|\vec w|}.$$

Here $$\vec u=\vec v$$ and $$\vec w=\vec a$$, so we require

$$\frac{|\vec v\cdot\vec a|}{|\vec a|}=19.$$

First we compute $$|\vec a|$$:

$$|\vec a|=\sqrt{2^{2}+(-1)^{2}+2^{2}}=\sqrt{4+1+4}=3.$$

Therefore the above condition becomes

$$|\vec v\cdot\vec a|=19\cdot3=57.$$

Now we evaluate $$\vec v\cdot\vec a$$ using $$\vec v=p\vec a+q\vec b$$:

$$\vec v\cdot\vec a=(p\vec a+q\vec b)\cdot\vec a=p(\vec a\cdot\vec a)+q(\vec b\cdot\vec a).$$

We already know $$\vec a\cdot\vec a=|\vec a|^{2}=9.$$ Next, calculate $$\vec b\cdot\vec a$$:

$$\vec b\cdot\vec a=2\cdot1+(-1)\cdot2+2\cdot(-1)=2-2-2=-2.$$

Hence

$$\vec v\cdot\vec a=9p-2q.$$

We therefore require

$$|9p-2q|=57.$$

But we already have the relation $$p=-4q$$ from the perpendicularity condition. Substituting $$p=-4q$$ gives

$$|9(-4q)-2q|=| -36q-2q |=|-38q|=38|q|=57.$$

Thus

$$|q|=\frac{57}{38}=\frac{3}{2}.$$

To proceed it is enough to take $$q=\pm\frac{3}{2}$$; whichever sign we choose, the square of the final magnitude will be the same. Correspondingly,

$$p=-4q=\mp6.$$

Now, to find $$|\vec v|^{2}$$ we use the expression

$$|\vec v|^{2}=(p\vec a+q\vec b)\cdot(p\vec a+q\vec b) = p^{2}(\vec a\cdot\vec a)+q^{2}(\vec b\cdot\vec b)+2pq(\vec a\cdot\vec b).$$

We already have $$\vec a\cdot\vec a=9$$ and $$\vec a\cdot\vec b=-2.$$ Next we compute $$\vec b\cdot\vec b$$:

$$\vec b\cdot\vec b=1^{2}+2^{2}+(-1)^{2}=1+4+1=6.$$

Substituting all these scalar products, we get

$$|\vec v|^{2}=9p^{2}+6q^{2}+2pq(-2)=9p^{2}+6q^{2}-4pq.$$

Using $$p=-4q$$, we substitute term by term:

$$p^{2}=16q^{2},\qquad pq=-4q^{2}.$$

Hence

$$|\vec v|^{2}=9(16q^{2})+6q^{2}-4(-4q^{2}) =144q^{2}+6q^{2}+16q^{2} =166q^{2}.$$

Now put $$q^{2}=\left(\frac{3}{2}\right)^{2}=\frac{9}{4},$$ obtaining

$$|\vec v|^{2}=166\cdot\frac{9}{4}=\frac{1494}{4}.$$

The question finally asks for $$|2\vec v|^{2}$$. Because multiplying a vector by the scalar $$2$$ multiplies its magnitude by $$2$$, we have

$$|2\vec v|^{2}=4|\vec v|^{2}=4\cdot\frac{1494}{4}=1494.$$

Hence, the correct answer is Option A.