Let $$\vec{u}$$ be a vector coplanar with the vectors $$\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$$ and $$\vec{b} = \hat{j} + \hat{k}$$. If $$\vec{u}$$ is perpendicular to $$\vec{a}$$ and $$\vec{u} \cdot \vec{b} = 24$$, then $$|\vec{u}|^2$$ is equal to:

We are told that the required vector $$\vec u$$ lies in the plane of the two given vectors, so the very first thing we write is that any vector coplanar with $$\vec a$$ and $$\vec b$$ can be expressed as their linear combination:

$$\vec u = p\,\vec a + q\,\vec b,$$

where $$p$$ and $$q$$ are real scalars to be determined.

The components of the given vectors are

$$\vec a = 2\hat i + 3\hat j - \hat k \; \Longrightarrow \; (2,\;3,\;-1),$$ $$\vec b = 0\hat i + 1\hat j + 1\hat k \; \Longrightarrow \; (0,\;1,\;1).$$

From these we calculate three basic scalar products that will repeatedly occur:

Magnitude‐square of $$\vec a$$: $$\vec a \cdot \vec a = 2^2 + 3^2 + (-1)^2 = 4 + 9 + 1 = 14.$$

Magnitude‐square of $$\vec b$$: $$\vec b \cdot \vec b = 0^2 + 1^2 + 1^2 = 0 + 1 + 1 = 2.$$

Mixed dot product: $$\vec a \cdot \vec b = 2\cdot 0 + 3\cdot 1 + (-1)\cdot 1 = 0 + 3 - 1 = 2.$$

Next, the condition “$$\vec u$$ is perpendicular to $$\vec a$$” translates to the dot product being zero:

$$\vec u \cdot \vec a = 0.$$

Substituting $$\vec u = p\vec a + q\vec b,$$ we get

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

Using the already computed dot products, this becomes

$$p(14) + q(2) = 0.$$

So,

$$14p + 2q = 0 \quad\Longrightarrow\quad q = -7p.$$

The problem also tells us that

$$\vec u \cdot \vec b = 24.$$

Again substituting $$\vec u = p\vec a + q\vec b,$$ we write

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

Putting the known numbers, we obtain

$$p(2) + q(2) = 24.$$

Dividing by 2 on both sides simplifies the expression:

$$p + q = 12.$$

Now we substitute $$q = -7p$$ (found earlier) into this linear equation:

$$p + (-7p) = 12 \quad\Longrightarrow\quad -6p = 12.$$

Hence,

$$p = -2,$$

and therefore

$$q = -7(-2) = 14.$$

With $$p$$ and $$q$$ known, we can finally work out the squared magnitude of $$\vec u$$. A general algebraic identity for a linear combination states

$$|\vec u|^2 \;=\; \vec u\cdot\vec u = (p\vec a + q\vec b)\cdot(p\vec a + q\vec b)$$ $$= p^2(\vec a\cdot\vec a) + q^2(\vec b\cdot\vec b) + 2pq(\vec a\cdot\vec b).$$

Putting the numerical values step by step:

$$p^2(\vec a\cdot\vec a) = (-2)^2 \times 14 = 4 \times 14 = 56,$$

$$q^2(\vec b\cdot\vec b) = 14^2 \times 2 = 196 \times 2 = 392,$$

$$2pq(\vec a\cdot\vec b) = 2 \times (-2) \times 14 \times 2 = -4 \times 14 \times 2 = -56 \times 2 = -112.$$

Adding these three contributions:

$$|\vec u|^2 = 56 + 392 - 112 = 336.$$

Hence, the correct answer is Option B.